\section{Introduction and main results}

The theory of tent spaces introduced by Coifman, Meyer and Stein~\cite{CMS}, and further studied by Cohn and Verbitsky~\cite{CohnVerbitsky} among others, shows the importance of maximal and square area functions and other objects from harmonic analysis~\cite{FefSt} in the study of Hardy spaces in the unit disc $\mathbb{D}=\{z:|z|<1\}$~\cite{DurenHp,Garnett1981}. The recent studies~\cite{PelRat,JAJRTent} show that tent spaces have natural analogues for Bergman spaces, and
they may play a role in the theory of weighted Bergman spaces similar to that of the original tent spaces in the Hardy space case.
The tent space $T^q_s(\nu,\om)$ consists of $\nu$-equivalence classes of
$\nu$-measurable functions $f:\mathbb{D}\to{\mathbb C}$ such that
    \begin{equation*}
    \begin{split}
    \|f\|_{T^q_s(\nu,\om)}^q&=\|A_{s,\nu}(f)\|_{L^q_\om}^q
    =\int_\mathbb{D}\left(\int_{\Gamma(\z)}|f(z)|^s\,d\nu(z)\right)^\frac{q}s\om(\z)\,dA(\z)
    <\infty,\quad0<q,s<\infty.
    \end{split}
    \end{equation*}
Here $\nu$ is assumed to be a positive Borel measure on $\mathbb{D}$, finite on compact sets, and $\om\in\widehat{\mathcal{D}}$, that is, $\om$ is radial and $\widehat{\om}(r)=\int_r^1\om(s)\,ds$ has the doubling property $\sup_{0\le r<1}\widehat{\om}(r)/\widehat{\om}(\frac{1+r}{2})<\infty$.
Moreover,
    $$
    \Gamma(z)=\left \{  \xi \in \mathbb{D} : |\theta- \arg(\xi)|<\frac{1}{2}\left(1-\frac{|\xi|}{r}\right )\right  \}, \quad z=re^{i\theta}\in\overline{\mathbb{D}}\setminus\{0\},
    $$
are non-tangential approach regions with vertexes inside the disc, and the related tents are defined by $T(\z)=\left\{z\in\mathbb{D}:\z\in\Gamma(z)\right\}$ for all $\z\in\mathbb{D}\setminus\{0\}$. We also set $\om(T(0))=\lim_{r\to0^+}\om(T(r))$ to deal with the origin.


The purpose of this paper is three fold. First, we are interested in the question of when the weighted Bergman space $A^p_\om$, consisting of analytic functions in the unit disc $\mathbb{D}$ such that
    $$
    \|f\|_{A^p_\om}^p=\int_\mathbb{D}|f(z)|^p\om(z)\,dA(z)<\infty,\quad 0<p<\infty,
    $$
is continuously or compactly embedded into the tent space $T^q_s(\nu,\om)$.
Analogous problems for Hardy and Hardy-Sobolev spaces have been considered in~\cite{CascanteOrtega2003,Cohn,GonLouWu2010}.
It turns out that the containment $A^p_\om\subset T^q_s(\nu,\om)$
is naturally described in terms of Carleson measures for $A^p_\om$. For $0<p,q<\infty$, a positive Borel measure $\mu$ on $\mathbb{D}$ is a $q$-Carleson measure for $A^p_\omega$ if there exists a constant $C>0$ such that $\|f\|_{L^q_\mu}\le C \| f \|_{A^p_\omega}$ for all $f\in A^p_\om$.


\begin{theorem}\label{Theorem1-intro}
Let $0<p,q,s<\infty$ such that $1+\frac{s}{p}-\frac{s}{q}>0$, $\om\in\widehat{\mathcal{D}}$ and $\nu$ a positive Borel measure on $\mathbb{D}$, finite on compact sets, such that $\nu(\{0\})=0$. Write $\nu_\om(\z)=\om(T(\z))\,d\nu(\z)$ for all $\z\in\mathbb{D}$. Then the following assertions hold:
    \begin{itemize}
    \item[\rm(i)] $I_d:A^p_\omega\to T^q_s(\nu,\omega)$ is bounded if and only if $\nu_\om$ is a $(p+s-\frac{ps}{q})$-Carleson measure for~$A^p_\om$. Moreover,
    $$
    \|I_d\|^s_{A^p_\omega\to T^q_s(\nu,\omega)}\asymp\|I_d\|^{p+s-\frac{ps}{q}}_{A^p_\omega\to L^{p+s-\frac{ps}{q}}_{\nu_\om}}.
    $$
    \item[\rm(ii)] $I_d:A^p_\omega\to T^q_s(\nu,\omega)$ is compact if and only if $I_d:A^p_\omega\to L^{p+s-\frac{ps}{q}}_{\nu_\om}$ is compact.
    \end{itemize}
\end{theorem}

The requirement $\nu(\{0\})=0$, which does not carry any real restriction, is a technical hypotheses caused by the geometry of the tents $\Gamma(z)$.

Theorem~\ref{Theorem1-intro} can be interpret as a characterization of Carleson measures. This is the second aim of our study and becomes more apparent when an operator is extracted from Theorem~\ref{Theorem1-intro}. For $0<s<\infty$, the generalized area operator induced by positive measures $\mu$ and $v$  on $\mathbb{D}$
is defined by
    $$
    G^v_{\mu,s}(f)(z)=\left(\int_{\Gamma(z)}|f(\z)|^s\frac{d\mu(\z)}{v(T(\z))}\right)^\frac1s,\quad z\in\mathbb{D}\setminus\{0\}.
    $$
Minkowski's inequality shows that $G_{\mu,s}^v$ is sublinear if $s\ge1$. This not the case for $0<s<1$, but instead we have
    $
    \left(G_{\mu,s}^v(f+g)\right)^s\le\left(G_{\mu,s}^v(f)\right)^s+\left(G_{\mu,s}^v(g)\right)^s
    $.
Anyway, we say that $G_{\mu,s}^v:A^p_\omega\to L^q_\omega$ is bounded if there exists $C>0$ such that $\|G_{\mu,s}^v(f)\|_{L^q_\omega}\le C\|f\|_{A^p_\omega}$ for all $f\in A^p_\omega$. Write $\mu^{\om}_v$ for the positive measure such that
    $$
    d\mu^{\om}_v(z)=\frac{\om(T(z))}{v(T(z))}\,d\mu(z)
    $$
for $\mu$-almost every $z\in\mathbb{D}$. Fubini's theorem shows that
    \begin{equation}\label{Eq:case-q=s}
    \begin{split}
    \|G_{\mu,s}^v (f)\|^s_{L^{s}_\omega}
    &=\int_\mathbb{D}\left(\int_{\Gamma(z)}|f(\z)|^s\frac{d\mu (\z)}{v(T(\z))}\right)\omega(z)\,dA(z)\\
    &=\int_\mathbb{D}\left(\int_{\Gamma(z)}|f(\z)|^s\frac{d\mu^{\om}_v (\z)}{\om(T(\z))}\right)\omega(z)\,dA(z)\\
    &=\int_{\mathbb{D}\setminus\{0\}}|f(\z)|^s\left(\frac{1}{\omega(T(\z))} \int_{T(\z)}\omega(z)\,dA(z)\right)d\mu^{\om}_v(\z)\\
    &=\int_{\mathbb{D}\setminus\{0\}}|f(\z)|^s\,d\mu^{\om}_v(\z)
    =\|f\|^s_{L^{s}_{\mu^{\om}_v}}-|f(0)|^s\mu^{\om}_v(\{0\}),
    \end{split}
    \end{equation}
and hence $G_{\mu,s}^v:A^p_\omega\to L^s_\omega$ is bounded if and only if $\mu^{\om}_v$ is an $s$-Carleson measure for $A^p_\omega$.
For any $s>0$, we say that $G_{\mu,s}^v: A^p_\om\to L^q_\mu$ is compact if for
every bounded sequence $\{f_n\}$ in $A^p_\om$ there exists a subsequence
$\{f_{n_k}\}$ such that $G_{\mu,s}^v(f_{n_k})$ converges in $L^q(\mu)$.

The next theorem gives a characterization of Carleson measures for Bergman spaces by using the generalized area operator $G_{\mu,s}^v$. Theorem~\ref{Theorem1-intro} is a special case of this result, see also Theorem~\ref{Theorem:main-s} in Section~\ref{Sec:boundedoperators}.

\begin{theorem}\label{Theorem2-intro}
Let $0<p,q,s<\infty$ such that $s>q-p$, $\omega\in\widehat{\mathcal{D}}$ and let $\mu,v$ be positive Borel measures on $\mathbb{D}$
such that $\mu\left(\{z\in\mathbb{D}:v(T(z))=0\}\right)=0=\mu(\{0\})$. Then the following assertions hold:
    \begin{itemize}
    \item[\rm(i)]$\mu^{\om}_v$ is a $q$-Carleson measure for~$A^p_\omega$ if and only if $G_{\mu,s}^v:A^p_\omega\to L^\frac{ps}{p+s-q}_\omega$ is bounded. Moreover,
    $$
    \|G_{\mu,s}^v\|^s_{A^p_\om\to L^\frac{ps}{p+s-q}_\om}\asymp\|I_d\|^q_{A^p_\omega\to L^q_{\mu^{\om}_v}}.
    $$
    \item[\rm(ii)] $I_d:A^p_\omega\to L^q_{\mu^{\om}_v}$ is compact if and only if $G_{\mu,s}^v:A^p_\omega\to L^\frac{ps}{p+s-q}_\omega$ is compact.
    \end{itemize}
\end{theorem}

The third motivation of this study comes from the equivalent $A^p_\om$-norm involving square functions, given by
    \begin{equation*}
    \begin{split}
    \|f\|_{A^p_\omega}^p\asymp\int_\mathbb{D}\,\left(\int_{\Gamma(\z)}|D^\alpha f(z)|^2
    \left(1-\left|\frac{z}{\z}\right|\right)^{2\alpha-2}\,dA(z)\right)^{\frac{p}2}\omega(\z)\,dA(\z)+\sum_{j=0}^{[\alpha]-1}|f^{(j)}(0)|^p.
    \end{split}
    \end{equation*}
Here $0<\alpha,p<\infty$, $\om$ is a radial weight, $D^\alpha f$ denotes the fractional derivative of order $\alpha$ and $[\alpha]$ is the integer such that $[\alpha]\le \alpha<[\alpha]+1$~\cite[Theorem $4.2$]{PelRat}.
This comparability shows that the operator
    $$
    F_\alpha(f)(\z)=\left(\int_{\Gamma(\z)}|D^\alpha f(z)|^2
    \left(1-\left|\frac{z}{\z}\right|\right)^{2\alpha-2}\,dA(z)\right)^{\frac{1}2},\quad
    \z\in\mathbb{D}\setminus\{0\},
    $$
is bounded from $A^p_\om$ to $L^p_\om$ for each $\alpha>0$. This is no longer true when $\alpha=0$, and therefore the definition of $G_{\mu,s}^v$ is also motivated by the study of this limit case. This was the starting point in the study by Cohn on the area operator $G_{\mu}(f)(z)=\int_{\Gamma(z)}|f(\z)|\frac{d\mu (\z)}{1-|\z|}$, defined for $z$ on the boundary $\mathbb{T}$ of $\mathbb{D}$, acting from the Hardy space $H^p$ to $L^p(\mathbb{T})$~\cite[Theorem~1]{Cohn}. The approach by Cohn relies on ideas by John and Nirenberg~\cite[Theorem~2.1]{Garnett1981} and Calderon-Zygmund decompositions. More recently, similar ideas together the classical factorization of Hardy spaces $H^p=H^{p_1}\cdot H^{p_2}$, $p^{-1}=p_1^{-1}+ p_2^{-1}$, were used in \cite{GonLouWu2010} to study the case $G_{\mu}:H^p\to L^q(\mathbb{T})$. We do not employ these techniques in the proof of Theorem~\ref{Theorem2-intro} but instead we use a description
of the boundedness of a weighted maximal function of H\"ormander-type. To give the precise statement we need to introduce some notation. The Carleson square $S(I)$ based on an interval $I\subset\mathbb{T}$ is the set $S(I)=\{re^{it}\in\mathbb{D}:e^{it}\in I,\,
1-|I|\le r<1\}$, where $|E|$ denotes the Lebesgue measure of $E\subset\mathbb{T}$. We associate to each $a\in\mathbb{D}\setminus\{0\}$ the interval
$I_a=\{e^{i\t}:|\arg(a e^{-i\t})|\le\frac{1-|a|}{2}\}$, and denote $S(a)=S(I_a)$. For a positive Borel measure $\mu$ on~$\mathbb{D}$ and $\alpha}       \def\b{\beta}        \def\g{\gamma>0$, define the weighted maximal function
    $$
    M_{\om,\alpha}(\mu)(z)=\sup_{z\in S(a)}\frac{\mu(S(a))}{\left(\om\left(S(a)
    \right)\right)^\alpha}       \def\b{\beta}        \def\g{\gamma},\quad
    z\in\mathbb{D}.
    $$
In the case $\alpha}       \def\b{\beta}        \def\g{\gamma=1$ simply write $M_{\om}(\mu)$, and if $\mu$ is of the form $\vp\om\,dA$, then $M_{\om,\alpha}       \def\b{\beta}        \def\g{\gamma}(\mu)$ is the weighted maximal function  $M_{\om,\alpha}       \def\b{\beta}        \def\g{\gamma}(\vp)$ of $\vp$. The following result is \cite[Theorem~3]{JAJRTent}.

\begin{lettertheorem}\label{co:maxbou}
Let $0<p\le q<\infty$ and $0<\gamma<\infty$ such that $p\gamma>1$.
Let $\om\in\widehat{\mathcal{D}}$ and $\mu$ be a positive Borel measure on $\mathbb{D}$. Then
$[M_{\om}((\cdot)^{\frac{1}{\gamma}})]^{\gamma}:L^p_\omega\to
L^q(\mu)$ is bounded if and only if $M_{\om,q/p}(\mu)\in L^\infty$.
Moreover,
    $$
    \|[M_{\om}((\cdot)^{\frac{1}{\gamma}})]^{\gamma}\|^q_{\left(L^p_\om,L^q(\mu)\right)}\asymp\|M_{\om,q/p}(\mu)\|_{L^\infty}.
    $$
\end{lettertheorem}

Another maximal operator we will face is defined by $N(f)(z)=\sup_{\z\in\Gamma(z)}|f(\z)|$. It is known that $N:A^p_\om\to L^p_\om$ is bounded for each radial weight $\om$ and $\|N(f)\|_{L^p_\om}\asymp\|f\|_{A^p_\om}$ by \cite[Lemma~4.4]{PelRat}.

Theorems~\ref{Theorem1-intro} and~\ref{Theorem2-intro} would perhaps not be of much significance if we did not understand the $q$-Carleson measures for $A^p_\om$ sufficiently well. In fact, another important ingredient in the proof of our main results is the following refinement of the characterizations of Carleson measures for Bergman spaces given in \cite[Theorem~1]{JAJRTent}. The estimates for the norm of the identity operator $I_d: A^p_\om\to L^q_\mu$ are of special importance for us.

\begin{theorem}\label{Theorem:normscm}
Let $0<p,q<\infty$, $\omega\in\widehat{\mathcal{D}}$ and let $\mu$ be a positive Borel measure on $\mathbb{D}$.
Further, let $dh(z)=dA(z)/(1-|z|^2)^2$ denote the hyperbolic measure.
    \begin{itemize}
    \item[\rm(i)] If $p\le q$, then $\mu$ is a $q$-Carleson measure for $A^p_\om$ if and only if
    $\sup_{a\in\mathbb{D}}\frac{\mu(S(a))}{\omega(S(a))^{\frac{q}{p}}}<\infty$. Moreover,
    $$
    \|I_d\|^q_{A^p_\om\to L^q_\mu}\asymp \sup_{a\in\mathbb{D}}\frac{\mu(S(a))}{\omega(S(a))^{\frac{q}{p}}}.
    $$
    \item[\rm(ii)] If $p\le q$, then $I_d:A^p_\om\to L^q_\mu$ is compact if and only if
    \begin{equation}\label{c1}
    \lim_{|a|\to 1^-}\frac{\mu(S(a))}{\left(\om(S(a))\right)^{q/p}}=0.
    \end{equation}
    \item[\rm(iii)] If $q<p$, then the following conditions are equivalent:
\begin{enumerate}
\item[\rm(a)] $I_d:A^p_\om\to L^q_\mu$ is compact;
\item[\rm(b)] $I_d:A^p_\om\to L^q_\mu$ is bounded;
 \item[\rm(c)] The function
    $$
    B^\om_\mu(z)=\int_{\Gamma(z)}\frac{d\mu(\z)}{\om(T(\z))},\quad
    z\in\mathbb{D}\setminus\{0\},
    $$
belongs to $L^{\frac{p}{p-q}}_\om$;
\item[\rm(d)] For each fixed $r\in(0,1)$, the function
    $$
    \Phi_\mu^\om(z)=\Phi_{\mu,r}^\om(z)=\int_{\Gamma(z)}\frac{\mu\left(\Delta(\z,r)\right)}{\om(T(\z))}\,dh(\z),\quad z\in\mathbb{D}\setminus\{0\},
    $$
belongs to $L^{\frac{p}{p-q}}_\om$;
\item[\rm(e)] For each sufficiently large $\lambda=\lambda(\om)>1$, the function
    $$
    \Psi^\om_\mu(z)=\Psi^\om_{\mu,\lambda}(z)=\int_\mathbb{D} \left(\frac{1-|\z|}{|1-\overline{z}\z|}\right)^\lambda\frac{d\mu(\z)}{\om(T(\z))},\quad z\in\mathbb{D},
    $$
belongs to $L^{\frac{p}{p-q}}_\om$.
\end{enumerate}
Moreover,
    $$
    \|I_d\|^q_{A^p_\om\to L^q_\mu}
    \asymp\|M_\om(\mu)\|_{L^{\frac{p}{p-q}}_\om}
    \asymp\|B^\om_\mu\|_{L^{\frac{p}{p-q}}_\om}+\mu(\{0\})
    \asymp\|\Psi^\om_\mu\|_{L^{\frac{p}{p-q}}_\om}
    \asymp\|\Phi^\om_\mu\|_{L^{\frac{p}{p-q}}_\om}+\mu(\{0\}).
    $$
    \end{itemize}
\end{theorem}

An analogue of the above result for Hardy spaces is essentially known. It can be obtained by using \cite[Section~7]{Lu90}
and \cite{PelPAMS14}. Going further, the implications of the techniques used in this paper can be employed to extend the known results on the area operator on Hardy spaces \cite{Cohn,GonLouWu2010}  as well as to the study of the integral operator $T_g(f)(z)=\int_0^zf(\z)g'(\z)\,d\z$ acting on Hardy and Bergman spaces. These results are briefly discussed in Section~\ref{Sec:comments}.







\section{Carleson measures}

In this section we prove Theorem~\ref{Theorem:normscm}. For this aim we need some preliminary results and definitions.
The following lemma provides useful characterizations of weights in $\widehat{\mathcal{D}}$. For a proof, see \cite{PelSum14,PRAntequera}.

\begin{letterlemma}\label{le:1}
Let $\om$ be a radial weight. Then the following conditions are equivalent:
\begin{itemize}
\item[\rm(i)] $\om\in\widehat{\mathcal{D}}$;
\item[\rm(ii)] There exist $C=C(\om)>0$ and $\b=\b(\om)>0$ such that
    \begin{equation}\label{Eq:replacement-Lemma1.1}
    \begin{split}
    \widehat{\om}(r)\le C\left(\frac{1-r}{1-t}\right)^{\b}\widehat{\om}(t),\quad 0\le r\le t<1;
    \end{split}
    \end{equation}
\item[\rm(iii)] There exist $C=C(\om)>0$ and $\gamma=\gamma(\om)>0$ such that
    \begin{equation}\label{Eq:replacement-Lemma1.2}
    \begin{split}
    \int_0^t\left(\frac{1-t}{1-s}\right)^\g\om(s)\,ds
    \le C\widehat{\om}(t),\quad 0\le t<1;
    \end{split}
    \end{equation}
\item[\rm(iv)] There exists $\lambda=\lambda(\om)\ge0$ such that
    $$
    \int_\mathbb{D}\frac{\om(z)}{|1-\overline{\z}z|^{\lambda+1}}\,dA(z)\asymp\frac{\widehat{\om}(\zeta)}{(1-|\z|)^\lambda},\quad \z\in\mathbb{D};
    $$
\item[\rm(v)] The associated weight
    $$
    \omega^\star(z)=\int_{|z|}^1\omega(s)\log\frac{s}{|z|}s\,ds,\quad z\in\mathbb{D}\setminus\{0\}.
    $$
satisfies
    $$
    \om(S(z))\asymp\om(T(z))\asymp\om^\star(z),\quad |z|\to1^-.
    $$
\end{itemize}
\end{letterlemma}

If $\om\in\widehat{\mathcal{D}}$, then Lemma~\ref{le:1} shows that for each $a\in\mathbb{D}$ and $\gamma=\gamma(\om)>0$ large enough, the function
    $$
    F_{a,p}(z)=\left( \frac{1-|a|^2}{1-\overline{a}z}\right)^{\frac{\gamma+1}{p}},\quad z\in\mathbb{D},
    $$
belongs to $A^p_\omega$ and satisfies $\|F_{a,p}\|_{A^p_\omega}^p\asymp\omega(S(a))$ and $|F_{a,p}(z)|\asymp 1$ for all $z\in S(a)$. This family of test functions will be frequently used in the sequel.

Apart from the tent spaces $T^q_s(\nu,\om)$, $0<q,s<\infty$, defined in the introduction, we will need to consider the case $q=\infty$. For $0<s<\infty$, define
    $$
    C^s_{s,\nu}(f)(\z)=\sup_{a\in \Gamma(\z)}\frac{1}{\om(T(a))}\int_{T(a)}|f(z)|^s\om(T(z))\,d\nu(z).
    $$
A quasi-norm in the tent space $T^\infty_s(\nu,\om)$ is defined by $\|f\|_{T^\infty_s(\nu,\om)}=\|C_{s,\nu}(f)\|_{L^\infty}$.

The pseudohyperbolic distance from $z$ to $w$ is defined by $\varrho(z,w)=\left|\frac{1-w}{1-\bar{z}w}\right|$, and the pseudohyperbolic
disc of center $a\in\mathbb{D}$ and radius $r\in(0,1)$ is denoted by $\Delta(a,r)=\{z:\varrho(a,z)<r\}$. The Euclidean discs are denoted by $D(a,r)=\{z\in{\mathbb C}:|a-z|<r\}$. Recall that $Z=\{z_k\}_{k=0}^\infty\subset\mathbb{D}$ is called a
separated sequence if it
is separated in the pseudohyperbolic metric, it is an $\e$-net if $\mathbb{D}=\bigcup_{k=0}^\infty \Delta(z_k,\e)$, and finally it is a
$\delta$-lattice if it is a $5\delta$-net and separated with constant $\gamma=\delta/5$. If  we have a discrete measure $\nu=\sum_k \delta}       \def\De{{\Delta}}    \def\e{\varepsilon} \def\Denw{{\Delta}^{\om}_{n}_{z_k}$, where $\{z_k\}$ is a separated sequence, then we write $T^p_q(\nu,\om)=T^p_q(\{z_k\},\om)$.


Recall that $I_d:A^p_{\om}\to L^q(\mu)$ is compact if it
maps bounded sets of $A^p_\om$ to relatively compact (precompact) sets of
$L^q(\mu)$. Equivalently, $I_d:A^p_{\om}\to L^q(\mu)$ is compact if and only if for
every bounded sequence $\{f_n\}$ in $A^p_\om$ there exists a subsequence that converges in $L^q(\mu)$.

\medskip

\noindent\emph{Proof of} Theorem~\ref{Theorem:normscm}. (i). There are several ways to bound the operator norm of $I_d:A^p_\om\to L^q_\mu$ from above by the claimed supremum. See \cite[Theorem~9]{JAJRTent} or \cite[Theorem~3.3]{PelSum14}, and also \cite[(11)]{JAJRTent} for the particular case $q=p$. The lower bound is obtained by using test functions, for details, see either~\cite[Lemma~8]{JAJRTent} or \cite[Theorem~3.3]{PelSum14}.

(ii). This case can be done by following the proof of \cite[Theorem~2.1(ii)]{PelRat}, with Lemma~\ref{le:1} in hand.

(iii). We first show that
    \begin{equation*}\label{ineq}
    \|I_d\|^q_{A^p_\om\to L^q_\mu}
    \le\|B^\om_\mu\|_{L^{\frac{p}{p-q}}_\om} +\mu(\{0\})
    \asymp\|\Psi^\om_\mu\|_{L^{\frac{p}{p-q}}_\om}
    \asymp\|\Phi^\om_\mu\|_{L^{\frac{p}{p-q}}_\om}+\mu(\{0\})
    \lesssim\|M_\om(\mu)\|_{L^{\frac{p}{p-q}}_\om}
    \lesssim\|I_d\|^q_{A^p_\om\to L^q_\mu}.
    \end{equation*}
Fubini's theorem, H\"older's inequality and \cite[Lemma~4.4]{PelRat} yield
    \begin{equation*}
    \begin{split}
    \|f\|^q_{L^{q}_\mu}
    &=\int_\mathbb{D}\left(\int_{\Gamma(z)}|f(\z)|^q \frac{d\mu(\z)}{\omega(T(\z))}\right)\omega(z)\,dA(z)+\mu(\{0\})|f(0)|^q\\
    &\le\int_\mathbb{D} N(f)^q(z)\int_{\Gamma(z)}\frac{d\mu(\z)}{\omega(T(\z))}\omega(z)\,dA(z)+ \mu(\{0\})|f(0)|^q\\
    &\lesssim \|N(f)\|^q_{L^{p}_\omega}\|B_\mu^\omega \|_{L^{\frac{p}{p-q}}_\omega}+\mu(\{0\})\|f\|^q_{A^{p}_\omega}
    \asymp\|f\|^q_{A^{p}_\omega}\left(\|B_\mu^\omega\|_{L^{\frac{p}{p-q}}_\omega}+\mu(\{0\}\right),
    \end{split}
    \end{equation*}
and hence $\|I_d\|^q_{A^p_\omega\to L^q_\mu}\lesssim\|B_\mu^\omega\|_{L^{\frac{p}{p-q}}_\omega}+\mu(\{0\})$. Moreover, $\|B_\mu^\omega\|_{L^{\frac{p}{p-q}}_\omega}+\mu(\{0\})\asymp\|\Psi_\mu^\omega\|_{L^{\frac{p}{p-q}}_\omega}$ by \cite[Lemma~4]{JAJRTent}.

Now write $d\nu(z)=\frac{d\mu(z)}{\om(T(z))}$ and $h_\lambda(\z,z)=\left(\frac{1-|\z|}{|1-\overline{\z}z|}\right)^\lambda$ for short. Then Fubini's theorem, \cite[Lemma~4]{JAJRTent} and the fact $\omega(T(\z))\asymp\omega(T(u))$ and $h_\lambda(\z,z)\asymp h_\lambda(u,z)$ for $\z\in\Delta(u,r)$, $r\in(0,1)$, yield
    \begin{equation}
    \begin{split}\label{eq:desc}
    \|\Psi^\omega_\mu\|^\frac{p}{p-q}_{L^\frac{p}{p-q}_\omega}
    &=\int_\mathbb{D}\left(\int_\mathbb{D}\frac{h_\lambda(\z,z)}{|\Delta(\z,r)|}|\Delta(\z,r)|\,d\nu(\z)\right)^\frac{p}{p-q}\om(z)\,dA(z)\\
    &\asymp\int_\mathbb{D}\left(\int_\mathbb{D} h_\lambda(u,z)\nu(\Delta(u,r))\,dh(u)\right)^\frac{p}{p-q}\om(z)\,dA(z)\\
    &\asymp\int_\mathbb{D}\left(\int_{\Gamma(z)}\nu(\Delta(u,r))\,dh(u)\right)^\frac{p}{p-q}\om(z)\,dA(z)+\mu(\{0\})\\
    &\asymp\int_\mathbb{D}\left(\int_{\Gamma(z)}\frac{\mu(\Delta(u,r))}{\om(T(u))}\,dh(u)\right)^\frac{p}{p-q}\om(z)\,dA(z)+\mu(\{0\})\\
    &=\|\Phi_\mu^\om\|^\frac{p}{p-q}_{L^\frac{p}{p-q}_\om}+\mu(\{0\})=\|g\|^\frac{p}{p-q}_{T^\frac{p}{p-q}_{1}(h,\omega)}+\mu(\{0\}),
    \end{split}
    \end{equation}
where $g(u)=\frac{\mu(\Delta(u,r))}{\om(T(u))}$. Now \cite[Lemma~7]{JAJRTent} implies
    \begin{equation}\label{10}
    \|g\|_{T^\frac{p}{p-q}_{1}(h,\omega)}\asymp\|C_{1,h}(g)\|_{L^\frac{p}{p-q}_\omega}.
    \end{equation}
Fubini's theorem gives
    \begin{equation*}
    \begin{split}
    C_{1,h}(g)(\z)&=\sup_{a\in\Gamma(\z)}\frac{1}{\omega(T(a))}\int_{T(a)}\mu(\Delta(z,r))\,dh(z)\\
    &=\sup_{\z\in T(a)}\frac{1}{\omega(T(a))}\int_{\mathbb{D}}\left(\int_{T(a)\cap\Delta(u,r)}dh(z)\right)d\mu(u).
    \end{split}
    \end{equation*}
The points $u\in\mathbb{D}$ for which $T(a)\cap\Delta(u,r)\ne\emptyset$ are contained in some tent $T(a')$, where $\arg a'=\arg a$ and $1-|a'|\asymp1-|a|$, for all $a\in\mathbb{D}\setminus D(0,\r)$, where $\r=\r(r)\in(0,1)$. Therefore
    \begin{equation*}
    \begin{split}
    \frac{1}{\omega(T(a))}\int_\mathbb{D}\left(\int_{T(a)\cap\Delta(u,r)}\,dh(z)\right)d\mu(u)
    &=\frac{1}{\omega(T(a))}\int_{T(a')}\left(\int_{T(a)\cap\Delta(u,r)}\,dh(z)\right)d\mu(u)\\
    &\lesssim\frac{\mu(T(a'))}{\omega(T(a))}\asymp\frac{\mu(T(a'))}{\omega(T(a'))},\quad a\in\mathbb{D}\setminus D(0,\r),
    \end{split}
    \end{equation*}
and it follows that
\begin{equation}\label{11}
    C_{1,h}(g)(\z)\asymp\sup_{\z\in T(a)}\frac{\mu(T(a))}{\omega(T(a))}
    \lesssim M_\omega(\mu)(\z),\quad\z\in\mathbb{D}\setminus\{0\}.
    \end{equation}
By combining \eqref{eq:desc}, \eqref{10} and \eqref{11},
we deduce
    \begin{equation*}
    \begin{split}
    \|\Psi^\omega_\mu\|_{L^\frac{p}{p-q}_\omega}
    &\asymp\|\Phi_\mu^\om\|_{L^\frac{p}{p-q}_\om}+\mu(\{0\})=\|g\|_{T^\frac{p}{p-q}_{1}(h,\omega)}+\mu(\{0\})\\
    &\asymp\|C_{1,h}(g)\|_{L^\frac{p}{p-q}_\omega}+\mu(\{0\})
    \lesssim\|M_\omega(\mu)\|_{L^\frac{p}{p-q}_\om}.
    \end{split}
    \end{equation*}
It remains to prove $\|M_\om(\mu)\|_{L^{\frac{p}{p-q}}_\om}\lesssim \|I_d\|^q_{A^p_\om\to L^q_\mu}$. To do this, we will show that $M_\om(\mu)$ is pointwise equivalent to the sum of two dyadic maximal functions.

Let $I_{n,k+l}=\{e^{i\theta}:\frac{\pi(k+l)}{2^{n+2}}\le\theta<\frac{\pi (k+l+1)}{2^{n+2}}\}$, and denote $\Upsilon_l=\{I_{n,k+l}:n\in\mathbb{N}\cup\{0\}$,  $k=0,1,\ldots,2^{n+2}-1\}$ and $l\in\{0,\frac12\}$. Define the dyadic maximal functions
    \begin{equation*}
    \begin{split}
    \widetilde{M}^d_{\omega,l}(\mu)(z)=\max\left\{\sup_{z\in T(I),I\in \Upsilon_l }\frac{\mu(T(I))}{\omega(T(I))} , \frac{\mu(\mathbb{D})}{\omega(\mathbb{D})}\right\},\quad z\in\mathbb{D},\quad l\in\left\{0,\frac12\right\},
    \end{split}
    \end{equation*}
and set
    $$
    \widetilde{M}^d_\omega(\mu)(z)=\widetilde{M}^d_{\omega,0}(\mu)(z)+\widetilde{M}^d_{\omega,\frac{1}{2}}(\mu)(z),\quad z\in\mathbb{D}.
    $$
If $\om\in\widehat{\mathcal{D}}$, then $\widetilde{M}^d_\omega(\mu)(z)\lesssim M_\omega(\mu)(z)$ for all $z\in\mathbb{D}$ because $\sup_{I\subset \mathbb{T}}\frac{\omega(S(I))}{\omega(T(I))}<\infty$ by Lemma~\ref{le:1}. For the converse inequality, given $I\subset \mathbb{T}$ such that $z\in S(I)$ there exist intervals $I_{n,k},I_{n,k+1}$ (if $k=2^{n+2}$, take $I_{n,k}=I_{n,0}$) such that $|I_{n+1,0}|<|I|\le|I_{n,0}|$, $I_{n,k}\cap I\neq\emptyset$ and $I_{n,k-1}\cap I=\emptyset$. We may assume that $n\ge3$, for otherwise the inequality we are searching for is immediate. Then $I\subset I_{n,k}\cup I_{n,k+1}$, and there exists $I_{n-3,m}\in\Upsilon_0\cup \Upsilon_{\frac{1}{2}}$ such that $\bigcup_{i=k-2}^{k+3}I_{n,i}\subset I_{n-3,m}$ and $S(I)\subset T(I_{n-3,m})$. Therefore Lemma~\ref{le:1} yields
    \begin{equation*}
    \begin{split}
    \frac{\mu(S(I))}{\omega(S(I))}
    \le\frac{\mu(T(I_{n-3,m}))}{\omega(T(I))} \leq \frac{\mu(T(I_{n-3,m}))}{\omega(T(I_{n+1,0}))}
    \asymp\frac{\mu(T(I_{n-3,m}))}{\omega(T(I_{n-3,m}))}\le\widetilde{M}^d_\omega(\mu)(z),\quad z\in S(I).
    \end{split}
    \end{equation*}
It follows that
    $
    M_\omega(\mu)(z)\lesssim\widetilde{M}^d_\omega(\mu)(z)
    $,
for all $z\in\mathbb{D}$, and hence
    \begin{equation}\label{eq:pointmax}
    M_\omega(\mu)(z)\asymp\widetilde{M}^d_\omega(\mu)(z),\quad z\in\mathbb{D}.
    \end{equation}

To estimate the norm of $\widetilde{M}^d_\omega(\mu)$ upwards, let choose $\{z_k\}$ be a separated sequence and
define
    $$
    S_\lambda(f)(z)=\sum_k f(z_k)\left(\frac{1-|z_k|}{1-\overline{z}_kz}\right)^\lambda,\quad z\in\mathbb{D}.
    $$
By \cite[Lemma~6]{JAJRTent} there exists $\lambda=\lambda(\om)>1$ such that $S_\lambda:\,T^p_{2}(\{z_k\},\omega)\to A^p_\omega$ is bounded. By denoting $\{b_{k}\}=\{f(z_{k})\}$, this implies
    \begin{equation*}
    \begin{split}
    \int_{\mathbb{D}}\left|\sum_{k}b_{k}h_\lambda(z_{k},z)\right|^q\,d\mu(z)
    &=\|S_\lambda(f)\|^q_{L^q(\mu)}
    \lesssim\|I_d\|^q_{A^p_\omega\to L^q_\mu}\|S_\lambda\|^q_{T^p_{2}(\{z_k\},\omega)\to A^p_\omega}\|\{b_k\}\|^q_{T^p_{2}(\{z_k\},\omega)}.
    \end{split}
    \end{equation*}
By replacing $b_k$ by $r_k(t)b_k$, where $r_k$ denotes the $k$th Rademacher function, using the fact that $|h_\lambda(z_{k},z)|\gtrsim\chi_{T(z_{k})}(z)$ for $z\in T(z_{k})$, and applying Khinchine's inequality, we deduce
    \begin{equation}\label{2}
    \begin{split}
    \int_\mathbb{D} \left( \sum_k |b_k|^2\chi_{T(z_k)}(z)\right)^\frac{q}{2}d\mu(z)
    \lesssim\|I_d\|^q_{A^p_\omega\to L^q_\mu}\|S_\lambda\|^q_{T^p_{2}(\{z_k\},\omega)\to A^p_\omega}\|\{b_k\}\|^q_{T^p_{2}(\{z_k\},\omega)}.
    \end{split}
    \end{equation}

Let $l\in\{0,\frac12\}$ be fixed. For each $k\in\mathbb{Z}$, let $\mathcal{E}_k$ denote the collection of maximal dyadic tents $T\in\{T(I):I\in\Upsilon_l\}\cup\{\mathbb{D}\}$ with respect to inclusion such that $\mu(T)>2^k\omega(T)$, and let $E_k=\cup_{T\in\mathcal{E}_k}T$. Then $2^k<\widetilde{M}^d_{\omega,l}(\mu)(z)\le2^{k+1}$ for $z \in E_k\setminus E_{k+1}$. Let now $\{b_T\}$ be a sequence indexed by $T\in\mathcal{E}=\cup_{k}\mathcal{E}_k$. Assume for a moment that $\mu$ has compact support. Then $\{b_T\}$ is a finite sequence. For $T\in\mathcal{E}$, let $G(T)=T-\cup\{T'\in\mathcal{E}:T'\subsetneq T\}$, and hence $G(T)=T-\cup\{T'\in\mathcal{E}_{k+1}:T'\subsetneq T\}$ for $T\in \mathcal{E}_k$. If $T_1,T_2\in\mathcal{E}$ are different (either one is strictly included in the other or they are disjoint), the sets $G(T_1)$ and $G(T_2)$ are disjoint, and hence
    \begin{equation}\label{3}
    \begin{split}
    \left( \sum_{T\in \mathcal{E}} |b_T|^2\chi_{T}(z)\right)^\frac{q}{2}\geq \left( \sum_{T\in \mathcal{E}} |b_T|^2\chi_{G(T)}(z)\right)^\frac{q}{2} = \sum_{T\in \mathcal{E}} |b_T|^q\chi_{G(T)}(z).
    \end{split}
    \end{equation}
Index the tents in $\mathcal{E}$ according to which $\mathcal{E}_k$ with maximal index they belong to, by writing $\mathcal{E}_k\setminus\cup_{m<k}\mathcal{E}_m=\{T_j^k:j\in\mathbb{N}\}$. Further, denote $b_{k,j}=b_{T_j^k}$ and let $z_{k,j}$ denote the vertex of $T^k_j$ i.e. $T^k_j=T(z_{k,j})$ (with the convenience that the vertex of $\mathbb{D}$ is the origin). The estimates \eqref{2} and \eqref{3} yield
    \begin{equation}
    \begin{split}\label{eq:4}
    &\sum_{k,j}b^q_{k,j}\left(\mu(T_j^k)-\sum_{T_i^{k+1}\subset T_j^k}\mu(T_i^{k+1})\right)
    \lesssim\|I_d\|^q_{A^p_\omega\to L^q_\mu}
    \|\{b_{k,j}\}\|^q_{T^p_{2}(\{z_{k,j}\},\omega)}\\
    &\quad=\|I_d\|^q_{A^p_\omega\to L^q_\mu}
    \left(\int_\mathbb{D}\left(\sum_{k,j}|b_{k,j}|^2\chi_{T_j^k}(z)\right)^\frac{p}{2}\omega(z)\,dA(z)\right)^\frac{q}{p}.
    \end{split}
    \end{equation}
Write $r=\frac{p}{q}$ for short, and choose $b_{k,j}^q=2^{k(r'-1)}$ for each $k$ and $j$. Then, by using the inequality $2^k<\widetilde{M}^d_{\omega,l}(\mu)(z)\le2^{k+1}$ for $z \in E_k\setminus E_{k+1}$, the left hand side of \eqref{eq:4} can be estimated as
    \begin{equation*}
    \begin{split}
    &\sum_{k,j}b^q_{k,j}\left(\mu(T_j^k)-\sum_{T_i^{k+1}\subset T_j^k}\mu(T_i^{k+1})\right)
    =\sum_{k}2^{k(r'-1)}\mu(E_k)-\sum_{k}2^{k(r'-1)}\sum_j\mu(T_j^k\cap E_{k+1})\\
    &=\sum_{k} (2^{k(r'-1)}-2^{(k-1)(r'-1)})\mu(E_k)
    =\left(1-\frac{1}{2^{r'-1}}\right)\sum_{k}2^{kr'}2^{-k}\sum_j\frac{\mu(T_j^k)}{\omega(T_j^k)}\omega(T_j^k)\\
    &\gtrsim\sum_k2^{kr'}\omega(E_k)
    \gtrsim\sum_{k}\int_{E_k\setminus E_{k+1}}\left(\widetilde{M}^d_{\omega,l}(\mu)(z)\right)^{r'}\omega(z)\,dA(z)
    =\|\widetilde{M}^d_{\omega,l}(\mu)\|_{L^{r'}_\omega}^{r'},
    \end{split}
    \end{equation*}
while the integral on the right hand side of \eqref{eq:4} with the notation $\eta=2^{(r'-1)\frac{2}{q}}$ becomes
    \begin{equation*}
    \begin{split}
    &\int_\mathbb{D}\left(\sum_{k,j}|b_{k,j}|^2\chi_{T_j^k}(z)\right)^\frac{p}{2}\omega(z)\,dA(z)
    =\int_\mathbb{D}\left(\sum_{k,j}\eta^k\chi_{T_j^k}(z)\right)^\frac{p}{2}\omega(z)\,dA(z)\\
    &=\int_\mathbb{D}\left(\sum_{k}\eta^k\chi_{E_k}(z)\right)^\frac{p}{2}\omega(z)\,dA(z)
    =\int_\mathbb{D}\left(\frac{\eta}{\eta-1}\sum_{k}\left(\eta^k-\eta^{k-1}\right)\chi_{E_k}(z)\right)^\frac{p}{2}\omega(z)\,dA(z)\\
    &\asymp\int_\mathbb{D}\left(\sum_{k}\eta^k\left(\chi_{E_k}(z)-\chi_{E_{k+1}}(z)\right)\right)^\frac{p}{2}\omega(z)\,dA(z)
    =\int_\mathbb{D}\left(\sum_{k}\eta^k \chi_{E_k\setminus E_{k+1} }(z)\right)^\frac{p}{2}\omega(z)\,dA(z)\\
    &=\int_\mathbb{D}\sum_{k} \eta^\frac{kp}{2} \chi_{E_k\setminus E_{k+1}}(z)\omega(z)\,dA(z)
    =\int_\mathbb{D}\sum_{k} 2^{r'k} \chi_{E_k\setminus E_{k+1} }(z)\omega(z)\,dA(z)\\
    &\asymp\sum_{k}\int_{E_k\setminus E_{k+1}}\left(\widetilde{M}^d_{\omega,l}(\mu)(z)\right)^{r'}\omega(z)\,dA(z)
    =\|\widetilde{M}^d_{\omega,l}(\mu)\|_{L^{r'}_\omega}^{r'}.
    \end{split}
    \end{equation*}
 Consequently, $\|\widetilde{M}^d_{\omega,l}(\mu)\|_{L^{r'}_\omega}^{r'}\lesssim\|I_d\|^q_{A^p_\omega\to L^q_\mu}\|\widetilde{M}^d_{\omega,l}(\mu)\|_{L^{r'}_\omega}^{\frac{r'}{r}}$, and thus $\|\widetilde{M}^d_{\omega,l}(\mu)\|_{L^{\frac{p}{p-q}}_\omega}\lesssim\|I_d\|^q_{A^p_\omega\to L^q_\mu}$ because $r'=\left(\frac{p}{q}\right)'=\frac{p}{p-q}$. Since this is valid for $l\in\{0,\frac12\}$, using Minkowski's inequality and
 \eqref{eq:pointmax} we get
    \begin{equation*}
    \begin{split}
    \|M_\omega(\mu)\|_{L^\frac{p}{p-q}_\omega}
    \asymp\|\widetilde{M}^d_\omega(\mu)\|_{L^{\frac{p}{p-q}}_\omega}
    \le\|\widetilde{M}^d_{\omega,0}(\mu)\|_{L^{\frac{p}{p-q}}_\omega}
    +\|\widetilde{M}^d_{\omega,\frac{1}{2}}(\mu)\|_{L^{\frac{p}{p-q}}_\omega}
    \lesssim\|I_d\|^q_{A^p_\omega\to L^q_\mu}
    \end{split}
    \end{equation*}
for $\mu$ with compact support. If $\mu$ is positive, then the above estimate, applied to the compactly supported $\mu_r=\chi_{D(0,r)}\mu$, and the standard limiting argument with monotone convergence theorem gives $\|M_\omega(\mu)\|_{L^\frac{p}{p-q}_\omega}\lesssim\|I_d\|^q_{A^p_\omega\to L^q_\mu}$. Hence the claimed operator norm estimates are valid and, in particular, (b)-(e) are equivalent.

To complete the proof of (iii), it suffices to show that $I_d:A^p_\om\to L^q_\mu$ is compact if (e) is satisfied.
By the hypothesis (e), \eqref{eq:desc} and
the dominated convergence theorem,
    \begin{equation}
    \begin{split}\label{s4}
    0 &=\lim_{R\to 1^-}\int_\mathbb{D} \left(\int_{\{R<|z|<1\}} \left(\frac{1-|z|}{|1-\overline{\z}z|}\right)^{\lambda} \frac{d\mu(z)}{\om(T(z))}\right)^{\frac{p}{p-q}}\,\om(\z)\,dA(\z)\\
    &\gtrsim \lim_{R\to 1^-} \int_\mathbb{D} \left(\int_{\Gamma(\z)\setminus\overline{D(0,R)}} \frac{d\mu(z)}{\om(T(z))}\right)^{\frac{p}{p-q}}\,\om(\z)\,dA(\z).
    \end{split}
    \end{equation}
Let $\{f_n\}$ be a bounded sequence in $A^p_\om$. Then $\{f_n\}$ is locally bounded and thus constitutes a normal family. Hence we may extract  a subsequence $\{f_{n_k}\}$ that converges uniformly on compact subsets of $\mathbb{D}$ to $f\in A^p_\om$. Write $g_k=f_{n_k}-f$. For $\e>0$, by \eqref{s4}, there exists $R_0=R_0\in(0,1)$ such that
    $$
    \int_\mathbb{D}\left(\int_{\Gamma(\z)\setminus\overline{D(0,R_0)}}\frac{d\mu(z)}{\om(T(z))}\right)^{\frac{p}{p-q}}\,\om(\z)\,dA(\z)<\e^{\frac{p}{p-q}}.
    $$
By the uniform convergence, we may choose $k_0\in\mathbb{N}$ such that $|g_k(z)|<\e^{1/q}$ for all $k\ge k_0$ and $z\in\overline{D(0,R_0)}$. Then Fubini's theorem, H\"older's inequality and \cite[Lemma~4.4]{PelRat} yield
    \begin{equation*}
    \begin{split}
    \|g_k\|^q_{L^{q}_\mu}
    &=\int_{\overline{D(0,R_0)}}|g_k(\z)|^q\,d\mu(\z)+\int_{\mathbb{D}\setminus\overline{D(0,R_0)}}|g_k(\z)|^q\,d\mu(\z)\\
    &\le\e\mu(\mathbb{D})+\int_\mathbb{D}\left(\int_{\Gamma(z)\setminus\overline{D(0,R_0)}}|g_k(\z)|^q \frac{d\mu(\z)}{\omega(T(\z))}\right)\omega(z)\,dA(z)\\
    &\le\e\mu(\mathbb{D})+\int_\mathbb{D} N(g_k)^q(z)\left(\int_{\Gamma(z)\setminus\overline{D(0,R_0)}}\frac{d\mu(\z)}{\omega(T(\z))}\right)\,\omega(z)\,dA(z)\\
    &\le\e\mu(\mathbb{D})+\|N(g_k)\|^q_{L^{p}_\omega}\e
    \asymp\e\mu(\mathbb{D})+\|g_k\|^q_{A^{p}_\omega}\e\lesssim \e,
    \end{split}
    \end{equation*}
and thus $I_d: A^p_\om \to L^q_\mu$ is compact.\hfill$\Box$








\section{Bounded operators $G_{\mu,s}^v:A^p_\omega\to L^q_\omega$}\label{Sec:boundedoperators}

Theorem~\ref{Theorem2-intro} is equivalent to the following result.

\begin{theorem}\label{Theorem:main-s}
Let $0<p,q,s<\infty$ such that $1+\frac{s}p-\frac{s}{q}>0$, $\omega\in\widehat{\mathcal{D}}$ and let $\mu,v$ be  positive Borel measures on $\mathbb{D}$
such that $\mu\left(\{z\in\mathbb{D}: v(T(z))=0\}\right)=0=\mu(\{0\})$. Then the following assertions hold:
    \begin{itemize}
    \item[\rm(i)]$G_{\mu,s}^v:A^p_\omega\to L^q_\omega$ is bounded if and only if $\mu^{\om}_v$ is a $\left(p+s-\frac{ps}{q}\right)$-Carleson measure for~$A^p_\omega$. Moreover,
    $$
    \|G_{\mu,s}^v\|^s_{A^p_\om\to L^q_\om}\asymp\|I_d\|^{p+s-\frac{ps}{q}}_{A^p_\omega\to L^{p+s-\frac{ps}{q}}_{\mu^{\om}_v}}.
    $$
    \item[\rm(ii)] $G_{\mu,s}^v:A^p_\omega\to L^q_\omega$ is compact if and only if
$I_d:A^p_\omega\to L^{p+s-\frac{ps}{q}}_{\mu^{\om}_v}$ is compact.
    \end{itemize}
\end{theorem}

Theorem~\ref{Theorem:main-s}(i) will be proved in two parts. We first deal with the case $q\ge p$.

\begin{theorem}\label{th:casepleq}
Let $0<p\le q<\infty$, $0<s<\infty$ and $\omega\in\widehat{\mathcal{D}}$, and let $\mu,v$ be  positive Borel measures on $\mathbb{D}$
such that $\mu\left(\{z\in\mathbb{D}: v(T(z))=0\}\right)=0=\mu(\{0\})$. Then $G_{\mu,s}^v : A^p_\omega \to L^q_\omega$ is bounded if and only if $\mu^{\om}_v$ is a $\left(p+s-\frac{ps}{q}\right)$-Carleson measure for $A^p_\omega$. Moreover,
    \begin{equation*}
    \begin{split}
    \|G_{\mu,s}^v\|^s_{A^p_\omega\to L^{q}_\omega}
    \asymp\|I_d\|^{p+s-\frac{ps}{q}}_{A^p_\omega\to L^{p+s-\frac{ps}{q}}_{\mu^{\om}_v}}
    \asymp\sup_{a\in\mathbb{D}}\frac{\mu^{\om}_v(S(a))}{\omega(S(a))^{1+\frac{s}{p}-\frac{s}{q}}}.
    \end{split}
    \end{equation*}
\end{theorem}

\begin{proof}
Let first $q>s$. Assume that $G_{\mu,s}^v:A^p_\omega\to L^q_\omega$ is bounded. Let $a\in\mathbb{D}$ and choose $\gamma=\gamma(p,q,s)$ sufficiently large so that  $\|F_{a,p}\|_{A^p_\omega}^p\asymp\omega(S(a))$ and $\|F_{a,(\frac qs)'}\|_{A^{(\frac qs)'}_\omega}^{(\frac qs)'}\asymp\omega(S(a))$. Then Fubini's theorem and H\"older's inequality yield
    \begin{equation*}
    \begin{split}
    \mu^{\om}_v(S(a))&\asymp \int_{S(a)} |F_{a,p}(z)|^s\,d\mu^{\om}_v(z)\\
    &\asymp\int_{S(a)}|F_{a,p}(z)|^s\left(\frac{1}{v(T(z))}\int_{T(z)}|F_{a,(\frac qs)'}(\z)|\omega(\z)\,dA(\z)\right)d\mu(z)\\
    &\lesssim\int_\mathbb{D}|F_{a,(\frac qs)'}(\z)|\left(\int_{\Gamma(\z)}|F_{a,p}(z)|^s\frac{d\mu(z)}{v(T(z))}\right)\omega(\z)\,dA(\z)\\
    &\le\|F_{a,(\frac qs)'}\|_{A^{(\frac qs)'}_\omega}\|G_{\mu,s}^v (F_{a,p})\|^s_{L^{q}_\omega}
    \le\|F_{a,(\frac qs)'}\|_{A^{(\frac qs)'}_\omega}\|G_{\mu,s}^v\|^s_{A^p_\omega\to L^{q}_\omega}\|F_{a,p}\|^s_{A^p_\omega}\\
    &\asymp\|G_{\mu,s}^v\|^s_{A^p_\omega\to L^{q}_\omega}\omega(S(a))^{\frac{1}{(\frac qs)'}} \omega(S(a))^{\frac{s}{p}}
    \asymp\|G_{\mu,s}^v\|^s_{A^p_\omega\to L^{q}_\omega} \omega(S(a))^{1+\frac{s}{p}-\frac{s}{q}}.
    \end{split}
    \end{equation*}
Hence
    $
    \sup_{a\in\mathbb{D}}\frac{\mu^{\om}_v(S(a))}{\omega(S(a))^{1+\frac{s}{p}-\frac{s}{q}}}\lesssim \|G_{\mu,s}^v\|^s_{A^p_\omega\to L^{q}_\omega}
    $
and $\mu^{\om}_v$ is a $\left(p+s-\frac{ps}{q}\right)$-Carleson measure for $A^p_\omega$ by Theorem~\ref{Theorem:normscm}.

Conversely, let $q>s$ and $\mu^{\om}_v$ be a $\left(p+s-\frac{ps}{q}\right)$-Carleson measure for $A^p_\omega$. Write $t=t(p,q,s)=p+s-\frac{ps}{q}>s$ for short. Since $\frac{t}{p}=(\frac{t}{s})'/(\frac{q}{s})'$, \cite[Theorem~3]{JAJRTent} shows that $M_\omega:L^{(\frac{q}{s})'}_\omega \to L^{(\frac{t}{s})'}_{\mu^{\om}_v}$ is bounded with $\|M_\om\|^{(\frac{t}{s})'}_{L^{(\frac{q}{s})'}_\omega\to L^{(\frac{t}{s})'}_{\mu^{\om}_v}}\asymp \sup_{a\in\mathbb{D}}\frac{\mu^{\om}_v(S(a))}{\omega(S(a))^{t/p}}$.
 This together with Theorem~\ref{Theorem:normscm}, Fubini's theorem and H\"older's inequality give
    \begin{equation*}
    \begin{split}
    \|G_{\mu,s}^v(f)\|^s_{L^{q}_\omega}&=\sup_{\|h\|_{L^{(\frac{q}{s})'}_\omega}\le1}\int_\mathbb{D}|h(z)|
    \left(\int_{\Gamma(z)}|f(\z)|^s\frac{d\mu^{\om}_v (\z)}{\omega(T(\z))}\right)\omega(z)\,dA(z)\\
    &=\sup_{\| h\|_{L^{(\frac{q}{s})'}_\omega}\le1}\int_\mathbb{D}|f(\z)|^s\left(\frac{1}{\omega(T(\z))}\int_{T(\z)}|h(z)|\omega(z)\,dA(z)\right)d\mu^{\om}_v(\z)\\
    &\lesssim\sup_{\| h\|_{L^{(\frac{q}{s})'}_\omega}\le1}\int_\mathbb{D}|f(\z)|^sM_\omega(h)(\z)\,d\mu^{\om}_v(\z)
    \le\sup_{\|h\|_{L^{(\frac{q}{s})'}_\omega}\le1}\|f\|^s_{L^{t}_{\mu^{\om}_v}}\| M_\omega(h)\|_{L^{(\frac{t}{s})'}_{\mu^{\om}_v}}\\
    &\le\sup_{\|h\|_{L^{(\frac{q}{s})'}_\omega}\le1}\|I_d\|^s_{A^p_\omega\to L^{t}_{\mu^{\om}_v}}\|f\|^s_{A^{p}_\omega}
    \|M_\om\|_{L^{(\frac{q}{s})'}_\omega\to L^{(\frac{t}{s})'}_{\mu^{\om}_v}}\|h\|_{L^{(\frac{q}{s})'}_\omega}\\
    &\asymp\left(\sup_{a\in\mathbb{D}}\frac{\mu^{\om}_v(S(a))}{\omega(S(a))^{1+\frac{s}{p}-\frac{s}{q}}}\right)^{\frac{s}t+\frac1{(\frac{t}{s})'}}\|f\|^s_{A^{p}_\omega},
    \end{split}
    \end{equation*}
and hence $G_{\mu,s}^v:A^p_\omega\to L^q_\omega$ is bounded and $\|G_{\mu,s}^v(f)\|^s_{L^{q}_\omega}\lesssim \sup_{a\in\mathbb{D}}\frac{\mu^{\om}_v(S(a))}{\omega(S(a))^{1+\frac{s}{p}-\frac{s}{q}}}$.

Since the assertion is valid for $q=s$ by \eqref{Eq:case-q=s}, it remains to consider the case $q<s$. Let first $\mu^{\om}_v$ be a $\left(p+s-\frac{ps}{q}\right)$-Carleson measure for $A^p_\omega$, and let $0<x<s$. H\"older's inequality and Fubini's theorem yield
    \begin{equation*}
    \begin{split}
    \|G_{\mu,s}^v(f)\|^q_{L^{q}_\omega}
    &=\int_\mathbb{D}\left(\int_{\Gamma(z)}|f(\z)|^{x+s-x}\frac{d\mu^{\om}_v(\z)}{\omega(T(\z))}\right)^\frac{q}{s} \omega(z)\,dA(z)\\
    &\le\int_\mathbb{D} N(f)(z)^{\frac{qx}{s}}\left(\int_{\Gamma(z)}|f(\z)|^{s-x}\frac{d\mu^{\om}_v(\z)}{\omega(T(\z))}\right)^\frac{q}{s}\omega(z)\,dA(z)\\
    &\le\left(\int_\mathbb{D} N(f)(z)^{\frac{qx}{s-q}}\omega(z)\,dA(z)\right)^{1-\frac{q}{s}}
    \left(\int_\mathbb{D}\int_{\Gamma(z)}|f(\z)|^{s-x}\frac{d\mu^{\om}_v(\z)}{\omega(T(\z))}\omega(z)\,dA(z)\right)^\frac{q}{s}\\
    &\lesssim\|N(f)\|^{\frac{qx}{s}}_{L_\omega^{\frac{qx}{s-q}}}\left(\int_\mathbb{D}|f(\z)|^{s-x}\,d\mu^{\om}_v (\z)\right)^\frac{q}{s}
    \asymp\|N(f)\|^{\frac{qx}{s}}_{L_\omega^{\frac{qx}{s-q}}}\|f\|^{\frac{q(s-x)}{s}}_{L_{\mu^{\om}_v}^{(s-x)}}.
    \end{split}
    \end{equation*}
Take $x=\frac{p(s-q)}{q}<s$ so that $s-x=s+p-\frac{ps}{q}$. Then the estimates above together with \cite[Lemma 4.4]{PelRat} and Theorem~\ref{Theorem:normscm} give
    \begin{equation*}
    \begin{split}
    \|G_{\mu,s}^v(f)\|^q_{L^{q}_\omega}
    \lesssim\|f\|^{\frac{qx}{s}}_{A_\omega^{p}}\|I_d\|^{\frac{q(s-x)}{s}}_{A_\omega^{p}\to L_{\mu^{\om}_v}^{s+p-\frac{ps}{q}}}\|f\|^{\frac{q(s-x)}{s}}_{A_\omega^{p}}
    \asymp\left(\sup_{a\in\mathbb{D}}\frac{\mu^{\om}_v(S(a))}{\omega(S(a))^{1+\frac{s}{p}-\frac{s}{q}}}\right)^\frac{q}{s}\|f\|^{q}_{A_\omega^{p}},
    \end{split}
    \end{equation*}
and hence $G_{\mu,s}^v:A^p_\omega\to L^q_\omega$ is bounded with $\|G_\mu^v\|^s_{A^p_\omega\to L^{q}_\omega}\lesssim \sup_{a\in\mathbb{D}}\frac{\mu^{\om}_v(S(a))}{\omega(S(a))^{1+\frac{s}{p}-\frac{s}{q}}}$.

Conversely, let $q<s$ and $G_{\mu,s}^v:A^p_\omega\to L^q_\omega$ be bounded. Choose $\alpha>\beta>1$ such that $\frac{\beta}{\alpha}=\frac{q}{s}$. Fubini's theorem and H\"older's inequality  yield
    \begin{equation}\label{1-s}
    \begin{split}
    \mu^{\om}_v(S(a))&=\int_\mathbb{D}\chi_{S(a)}(z)\,d\mu^{\om}_v(z)
    \asymp\int_\mathbb{D}\chi_{S(a)}(z)|F_{a,p}(z)|^s\frac{1}{v(T(z))}\int_{T(z)}\omega(\z)\,dA(\z)\,d\mu(z)\\
    &=\int_\mathbb{D}\left(\int_{\Gamma(\z)}\chi_{S(a)}(z)|F_{a,p}(z)|^s\frac{ d\mu (z)}{v(T(z))}\right)^{\frac{1}{\alpha}+\frac{1}{\alpha'}}\omega(\z)\,dA(\z)\\
    &\le\left(\int_\mathbb{D} \left(\int_{\Gamma(\z)}|F_{a,p}(z)|^s\frac{d\mu(z)}{v(T(z))}\right)^{\frac{\beta}{\alpha}}\omega(\z)\,dA(\z)\right)^{\frac{1}{\beta}}\\
    &\quad\cdot\left(\int_\mathbb{D}\left(\int_{\Gamma(\z)}\chi_{S(a)}(z)\frac{d\mu(z)}{v(T(z))}\right)^{\frac{\beta'}{\alpha'}}
    \omega(\z)\,dA(\z)\right)^{\frac{1}{\beta'}} \\
    &=\|G_{\mu,s}^v(F_{a,p})\|^{\frac{q}{\beta}}_{L^{q}_\omega}\|G_{\mu,1}^v(\chi_{S(a)})\|^{\frac{1}{\alpha'}}_{L^{\frac{\beta'}{\alpha'}}_\omega}.
    \end{split}
    \end{equation}
Now $\frac{\beta'}{\alpha'}>1$ because $\beta<\alpha$, and hence $\|G_{\mu,1}^v( \chi_{S(a)})\|_{L^{\frac{\beta'}{\alpha'}}_\omega}$ can be estimated by duality arguments. Namely, since $\left(\frac{\beta'}{\alpha'}\right)'=\frac{\beta(\alpha-1)}{\alpha-\beta}$, Fubini's theorem, H\"older's inequality and \cite[Theorem~3]{JAJRTent} give
    \begin{equation}\label{1-s'}
    \begin{split}
    \|G_{\mu,1}^v(\chi_{S(a)})\|_{L^{\frac{\beta'}{\alpha'}}_\omega}
    &=\sup_{\|h\|_{L^{\frac{\beta(\alpha-1)}{\alpha-\beta}}_\omega}\le1}\int_\mathbb{D}|h(\z)|G_{\mu,1}^v(\chi_{S(a)})(\z) \omega(\z)\,dA(\z)\\
    &=\sup_{\|h\|_{L^{\frac{\beta(\alpha-1)}{\alpha-\beta}}_\omega}\le1}\int_\mathbb{D}|h(\z)|\left(\int_{\Gamma(\z)}\chi_{S(a)}(z)\frac{d\mu (z)}{v(T(z))} \right) \omega(\z)\,dA(\z)\\
    &=\sup_{\|h\|_{L^{\frac{\beta(\alpha-1)}{\alpha-\beta}}_\omega}\le1}\int_\mathbb{D}\chi_{S(a)}(z)
    \left(\frac{1}{v(T(z))}\int_{T(z)}|h(\z)|\omega(\z)\,dA(\z)\right)d\mu(z)\\
    &=\sup_{\|h\|_{L^{\frac{\beta(\alpha-1)}{\alpha-\beta}}_\omega}\le1}\int_\mathbb{D}\chi_{S(a)}(z)
    \left(\frac{1}{\om(T(z))}\int_{T(z)}|h(\z)|\omega(\z)\,dA(\z)\right)d\mu^{\om}_v(z)\\
    &\lesssim\sup_{\|h\|_{L^{\frac{\beta(\alpha-1)}{\alpha-\beta}}_\omega}\le1}\int_\mathbb{D}\chi_{S(a)}(z)M_\omega(h)(z)\,d\mu^{\om}_v(z)\\
    &\le\sup_{\|h\|_{L^{\frac{\beta(\alpha-1)}{\alpha-\beta}}_\omega}\le1}\left(\int_\mathbb{D}\chi_{S(a)}(z)\,d\mu^{\om}_v(z)\right)^{\frac{\alpha'}{\beta'}}\\
    &\quad\cdot\left(\int_\mathbb{D} M_\omega(h)(z)^\frac{\beta(\alpha-1)}{\alpha-\beta}\chi_{S(a)}(z)\,d\mu^{\om}_v(z)\right)^{1-\frac{\alpha'}{\beta'}}\\
    &\lesssim\mu^{\om}_v(S(a))^{\frac{\alpha'}{\beta'}} \left(\sup_{z\in \mathbb{D}} \frac{\mu^{\om}_v(S(z)\cap S(a))}{\omega(S(z))} \right)^{1-\frac{\alpha'}{\beta'}},\quad a\in\mathbb{D}.
    \end{split}
    \end{equation}
By combining this with \eqref{1-s} and using the norm estimate $\|F_{a,p}\|_{A^p_\omega}^p\asymp\omega(S(a))$, we deduce
    \begin{equation*}
    \begin{split}
    \mu^{\om}_v (S(a))
    &\lesssim\|G_{\mu,s}^v(F_{a,p})\|^{\frac{q}{\beta}}_{L^{q}_\omega}
    \|G_{\mu,1}^v(\chi_{S(a)})\|^{\frac{1}{\alpha'}}_{L^{\frac{\beta'}{\alpha'}}_\omega}\\
    &\lesssim\|G_{\mu,s}^v\|^{\frac{q}{\beta}}_{A^p_\omega\to L^{q}_\omega}\omega(S(a))^{\frac{q}{p\beta}} \mu^{\om}_v(S(a))^{\frac{1}{\beta'}} \left(\sup_{z\in \mathbb{D}} \frac{\mu^{\om}_v(S(z)\cap S(a))}{\omega(S(z))} \right)^{\frac{1}{\alpha'}-\frac{1}{\beta'}},
    \end{split}
    \end{equation*}
which yields
    \begin{equation}
    \begin{split}\label{eq:1-s}
    \mu^{\om}_v (S(a))^{\frac{1}{\beta}}\omega(S(a))^{-\frac{q}{p\b}}
    &\lesssim\|G_{\mu,s}^v\|^{\frac{q}{\b}}_{A^p_\omega\to L^{q}_\omega} \left(\sup_{z\in \mathbb{D}} \frac{\mu^{\om}_v(S(z)\cap S(a))}{\omega(S(z))} \right)^{\frac{1}{\alpha'}-\frac{1}{\beta'}},\quad a\in\mathbb{D}.
    \end{split}
    \end{equation}

Define $d\mu_r(z)=\chi_{D(0,r)}(z)d\mu(z)$ for $0<r<1$. Then
    \begin{equation}
    \begin{split}\label{eq:2-s}
    \left(G_{\mu_r,s}^v(f)(z)\right)^s
    &=\int_{\Gamma(z)}|f(\z)|^s\frac{d\mu_r (\z)}{v(T(\z))}
    =\int_{\Gamma(z)\cap D(0,r)}|f(\z)|^s\frac{d\mu (\z)}{v(T(\z))}\\
    &\le\int_{\Gamma(z)}|f(\z)|^s\frac{d\mu (\z)}{v(T(\z))}
    =\left(G_{\mu,s}^v(f)(z)\right)^s,\quad z\in\mathbb{D}\setminus\{0\},
    \end{split}
    \end{equation}
and hence $\|G_{\mu_r,s}^v\|_{A^p_\omega\to L^{q}_\omega}\le\|G_{\mu,s}^v\|_{A^p_\omega\to L^{q}_\omega}$ for all $0<r<1$.

If $q=p$, then \eqref{eq:1-s} applied to $\mu_r$ implies
    \begin{equation*}
    \begin{split}
    (\mu^{\om}_v)_r(S(a))^{\frac{1}{\beta}}\omega(S(a))^{-\frac{1}{\beta}}
    &\lesssim\|G_{\mu_r,s}^v\|^{\frac{p}{\beta}}_{A^p_\omega\to L^{p}_\omega}\left(\sup_{z\in\mathbb{D}}\frac{(\mu^{\om}_v)_r(S(z))}{\omega(S(z))} \right)^{\frac{1}{\alpha'}-\frac{1}{\beta'}},\quad a\in\mathbb{D},
    \end{split}
    \end{equation*}
and hence
    \begin{equation*}
    \begin{split}
    \left(\sup_{z\in\mathbb{D}}\frac{(\mu^{\om}_v)_r(S(z))}{\omega(S(z))} \right)^{\frac{1}{\beta}}
    &\lesssim\|G_{\mu,s}^v\|^{\frac{p}{\beta}}_{A^p_\omega\to L^{p}_\omega}\left(\sup_{z\in\mathbb{D}}\frac{(\mu^{\om}_v)_r(S(z))}{\omega(S(z))}\right)^{\frac{1}{\alpha'}-\frac{1}{\beta'}}.
    \end{split}
    \end{equation*}
Consequently,
    \begin{equation*}
    \begin{split}
    \sup_{z\in\mathbb{D}, r\in (0,1)}\frac{(\mu^{\om}_v)_r(S(z))}{\omega(S(z))}
    &\lesssim\|G_{\mu,s}^v\|^s_{A^p_\omega\to L^{p}_\omega}.
    \end{split}
    \end{equation*}
So Fatou's lemma and Theorem~\ref{Theorem:normscm} show that $\mu^{\om}_v$ is a $p$-Carleson measure for $A^p_\omega$ with
    $$
    \|I_d\|^p_{A^p_\om\to L^p_{\mu^{\om}_v}}
    \asymp\sup_{a\in\mathbb{D}}\frac{\mu^{\om}_v(S(a))}{\omega(S(a))}
    \lesssim\|G_{\mu,s}^v\|^s_{A^p_\omega\to L^{p}_\omega}.
    $$

If $q>p$, then applying \eqref{eq:1-s}  to $\mu_r$ and bearing in mind that $\om$ is radial
    \begin{equation*}
    \begin{split}
    \frac{(\mu^{\om}_v)_r(S(a))}{\omega(S(a))^{1+\frac{s}{p}-\frac{s}{q}}}
    &\lesssim\|G_{\mu_r,s}^v\|^q_{A^p_\omega\to L^{q}_\omega}\omega(S(a))^{\frac{q}{p}-1-\frac{s}{p}+\frac{s}{q}}
    \left(\sup_{z\in\mathbb{D}}\frac{(\mu^{\om}_v)_r(S(z)\cap S(a))}{\omega(S(z))}\right)^{\frac{\beta}{\alpha'}-\frac{\beta}{\beta'}}\\
    &=\|G_{\mu_r,s}^v\|^q_{A^p_\omega\to L^{q}_\omega}\omega(S(a))^{\frac{-(q-p)(s-q)}{pq}}
    \left(\sup_{z:S(z)\subset S(a)}\frac{(\mu^{\om}_v)_r(S(z)\cap S(a))}{\omega(S(z))}\right)^{1-\frac{q}{s}}\\
    &=\|G_{\mu_r,s}^v\|^q_{A^p_\omega\to L^{q}_\omega}
    \left(\sup_{z:S(z)\subset S(a)}\frac{(\mu^{\om}_v)_r(S(z)\cap S(a))}{\omega(S(a))^{s\frac{q-p}{pq}}\omega(S(z))}\right)^{1-\frac{q}{s}}\\
    &\le\|G_{\mu_r,s}^v\|^q_{A^p_\omega \to L^{q}_\omega}
    \left(\sup_{z:S(z)\subset S(a)}\frac{(\mu^{\om}_v)_r(S(z))}{\omega(S(z))^{1+\frac{s}{p}-\frac{s}{q}}}\right)^{1-\frac{q}{s}},\quad a\in\mathbb{D}.
    \end{split}
    \end{equation*}
Consequently,
    \begin{equation*}
    \begin{split}
    \sup_{a\in\mathbb{D}}\left(\frac{(\mu^{\om}_v)_r(S(a))}{\omega(S(a))^{1+\frac{s}{p}-\frac{s}{q}}}\right)
    &\lesssim\|G_{\mu,s}^v\|^q_{A^p_\omega\to L^{q}_\omega}
    \left(\sup_{a\in\mathbb{D}}\sup_{z:S(z)\subset S(a)}\frac{(\mu^{\om}_v)_r(S(z))}{\omega(S(z))^{1+\frac{s}{p}-\frac{s}{q}}}\right)^{1-\frac{q}{s}}\\
    &=\|G_{\mu,s}^v\|^q_{A^p_\omega\to L^{q}_\omega}\left(\sup_{a\in \mathbb{D}}\frac{(\mu^{\om}_v)_r(S(a))}{\omega(S(a))^{1+\frac{s}{p}-\frac{s}{q}}} \right)^{1-\frac{q}{s}},
    \end{split}
    \end{equation*}
and thus
    \begin{equation*}
    \begin{split}
    \sup_{a\in\mathbb{D},\,r\in(0,1)}\frac{(\mu^{\om}_v)_r(S(a))}{\omega(S(a))^{1+\frac{s}{p}-\frac{s}{q}}}
    &\lesssim\|G_{\mu,s}^v\|^s_{A^p_\omega\to L^{q}_\omega}.
    \end{split}
    \end{equation*}
Fatou's lemma and Theorem~\ref{Theorem:normscm} show that $\mu^{\om}_v$ is a $\left(p+s-\frac{ps}{q}\right)$-Carleson measure for~$A^p_\omega$ with the corresponding inequality of norms.
\end{proof}

\begin{theorem}\label{fukifuki}
Let $0<q<p<\infty$ and $0<s<\infty$ such that $1+\frac{s}p-\frac{s}{q}>0$, $\omega\in\widehat{\mathcal{D}}$ and let $\mu,v$ be  positive Borel measures on $\mathbb{D}$
such that $\mu\left(\{z\in\mathbb{D}: v(T(z))=0\}\right)=0=\mu(\{0\})$. Then $G_{\mu,s}^v:A^p_\omega\to L^q_\omega$ is bounded if and only if $\mu^{\om}_v$ is a $p\left(1+\frac{s}p-\frac{s}{q}\right)$-Carleson measure for $A^p_\omega$. Moreover,
    \begin{equation*}
    \begin{split}
    \|G_{\mu,s}^v\|^s_{A^p_\omega\to L^{q}_\omega}
   
    \asymp\|I_d\|^{p+s-\frac{ps}{q}}_{A^p_\omega\to L^{p+s-\frac{ps}{q}}_{\mu^{\om}_v}}
    \asymp\|B_\mu^v\|_{L^{\frac{qp}{s(p-q)}}_\omega}.
   
    \end{split}
    \end{equation*}
\end{theorem}

\begin{proof}
The equivalence $\|I_d\|^{p+s-\frac{ps}{q}}_{A^p_\omega\to L^{p+s-\frac{ps}{q}}_{\mu^{\om}_v}}
    \asymp\|B_\mu^v\|_{L^{\frac{qp}{s(p-q)}}_\omega}$ follows from Theorem~\ref{Theorem:normscm}.
\par If $B_\mu^v\in L^{\frac{qp}{s(p-q)}}_\omega$, then H\"older's inequality and \cite[Lemma~4.4]{PelRat} give
    \begin{equation*}
    \begin{split}
    \|G_{\mu,s}^v(f)\|^q_{L^{q}_\omega}
    &=\int_\mathbb{D}\left(\int_{\Gamma(z)}|f(\z)|^s\frac{d\mu(\z)}{v(T(\z))}\right)^\frac{q}{s}\omega(z)\,dA(z)\\
    &\le\int_\mathbb{D} N(f)^q(z)\left(\int_{\Gamma(z)}\frac{d\mu(\z)}{v(T(\z))}\right)^\frac{q}{s}\omega(z)\,dA(z)\\
    &\le\|N(f) \|^q_{L^{p}_\omega}\|B_\mu^v\|^\frac{q}{s}_{L^{\frac{qp}{s(p-q)}}_\omega}
    \asymp\|f\|^q_{A^{p}_\omega}\|B_\mu^v\|^\frac{q}{s}_{L^{\frac{qp}{s(p-q)}}_\omega},
    \end{split}
    \end{equation*}
and hence $G_{\mu,s}^v:A^p_\omega\to L^q_\omega$ is bounded and $\|G_{\mu,s}^v\|^s_{A^p_\omega\to L^{q}_\omega}\lesssim\|B_\mu^v\|_{L^{\frac{qp}{p-q}}_\omega}$.

Assume now that $G_{\mu,s}^v:A^p_\omega\to L^q_\omega$ is bounded, and let first $q>s$ and write $t=t(p,q,s)=(p+s-\frac{ps}{q})=s+p(1-\frac{s}{q})$. It suffices to show that $\mu^{\om}_v$ is a $t$-Carleson measure for $A^p_\omega$. Fubini's theorem, H\"older's inequality and \cite[Lemma~4.4]{PelRat} yield
    \begin{equation*}
    \begin{split}
    \|f\|_{L^t_{\mu^{\om}_v}}^t
    &=\int_\mathbb{D}|f(z)|^t\left(\int_{T(z)}\om(\z)\,dA(\z)\right)\frac{d\mu(z)}{v(T(z))}\\
    &=\int_\mathbb{D}\left(\int_{\Gamma(\z)}|f(z)|^{s+p(1-\frac{s}{q})}\frac{d\mu(z)}{v(T(z))}\right)\om(\z)\,dA(\z)\\
    &\le\int_\mathbb{D} N(f)^{p(1-\frac{s}{q})}(\z)\left(\int_{\Gamma(\z)}|f(z)|^s\frac{d\mu(z)}{v(T(z))}\right)\om(\z)\,dA(\z)\\
    &\lesssim\|N(f)\|_{L^p_\om}^{p(1-\frac{s}{q})}\|G^v_{\mu,s}(f)\|^s_{L^q_\om}
    \asymp\|f\|_{A^p_\om}^{p(1-\frac{s}{q})}\|G^v_{\mu,s}(f)\|^s_{L^q_\om}\\
    &\le \|G^v_\mu\|^s_{A^p_\om\to L^q_\om}\|f\|_{A^p_\om}^t
    \end{split}
    \end{equation*}
and hence $I_d:\,A^p_\om\to L^t_{\mu^{\om}_v}$ is bounded with
 $\|I_d\|^t_{A^p_\om\to L^t_{\mu^{\om}_v}}\lesssim\|G^v_\mu\|^s_{A^p_\om\to L^q_\om}$.

If $q=s$, then $t=s$ and the result follows from \eqref{Eq:case-q=s}.
\par If $q<s$, then $t=p+s-\frac{ps}{q}<s$ and $\frac{q}{t}>1$ because $\min\{p,s\}>q$. Hence H\"older's inequality gives
    \begin{equation*}
    \begin{split}
    \|f\|^t_{L^t_{\mu^{\om}_v}}
    &=\int_\mathbb{D}\left(\int_{\Gamma(\z)}|f(z)|^t\frac{d\mu(z)}{v(T(z))}\right)\omega(\z)\,dA(\z)\\
    &\lesssim\int_\mathbb{D}\left(\int_{\Gamma(\z)}|f(z)|^s\frac{d\mu(z)}{v(T(z))}\right)^\frac{t}{s}
    \left(\int_{\Gamma(\z)}\frac{d\mu(z)}{v(T(z))}\right)^{1-\frac{t}{s}}\omega(\z)\,dA(\z)\\
    &\le\left(\int_\mathbb{D}\left(\int_{\Gamma(\z)}|f(z)|^s\frac{d\mu(z)}{v(T(z))}\right)^\frac{q}{s}\omega(\z)\,dA(\z)\right)^\frac{t}{q} \\&\cdot\left(\int_\mathbb{D}\left(\int_{\Gamma(\z)}\frac{d\mu(z)}{v(T(z))}\right)^{\frac{s-t}{s}\frac{q}{q-t}}\omega(\z)\,dA(\z)\right)^\frac{q-t}{q}\\
    &=\|G_{\mu,s}^v(f)\|_{L^q_\omega}^t\|B_\mu^v\|_{L^\frac{qp}{s(p-q)}_\omega}^{1-\frac{t}{s}}\\
    &\le\|G_{\mu,s}^v\|_{A^p_\omega\to L^q_\omega}^t\|B_{\mu,s}^v\|_{L^\frac{qp}{s(p-q)}_\omega}^{1-\frac{t}{s}} \|f\|_{A^p_\omega}^t.
    \end{split}
    \end{equation*}
This applied to $\mu_r$ yields
    \begin{equation*}
    \begin{split}
    \left(\frac{\|f\|_{L^{t}_{(\mu^{\om}_v)_r}}}{\| f \|_{A^p_\omega}}\right)^t
    \lesssim\|G_{\mu_r,s}^v\|_{A^p_\omega\to L^q_\omega}^t
    \|B_{\mu_r}^v\|_{L^{\frac{qp}{s(p-q)}}_\omega}^{1-\frac{t}{s}}  ,\quad f\in A^p_\omega,\quad f\not\equiv0,
    \end{split}
    \end{equation*}
and hence
    \begin{equation*}
    \begin{split}
    \|I_d\|_{ A^p_\omega\to L^t_{{(\mu^{\om}_v)_r}}}^t
    \le\|G_{\mu_r,s}^v\|_{A^p_\omega\to L^q_\omega}^t\|B_{\mu_r}^v\|_{L^{\frac{qp}{s(p-q)}}_\omega}^{1-\frac{t}{s}}.
    \end{split}
    \end{equation*}
Since
    $
    \|I_d\|_{A^p_\omega\to L^t_{({\mu^{\om}_v})_r}}^t\asymp\|B_{\mu_r}^v\|_{L^\frac{qp}{s(p-q)}_\omega}
    $
by Theorem~\ref{Theorem:normscm}, we deduce
    \begin{equation*}
    \begin{split}
    \|B_{\mu_r}^v\|_{L^\frac{qp}{s(p-q)}_\omega}
    \lesssim\| G_{\mu_r,s}^v \|_{ A^p_\omega \to L^q_\omega}^t\|B_{\mu_r}^v\|_{L^{\frac{qp}{s(p-q)}}_\omega}^{1-\frac{t}{s}},
    \end{split}
    \end{equation*}
and hence
    \begin{equation*}
    \begin{split}
    \|B_{\mu_r}^v\|_{L^\frac{qp}{s(p-q)}_\omega}
    \lesssim\|G_{\mu_r,s}^v\|^s_{A^p_\omega\to L^q_\omega}
    \le\|G_{\mu,s}^v\|^s_{A^p_\omega\to L^q_\omega},
    \end{split}
    \end{equation*}
which together with Fatou's lemma gives
    $$
    \|B_{\mu}^v\|_{L^\frac{qp}{s(p-q)}_\omega}
    \le\liminf_{r\to 1}\|B_{\mu_r}^v\|_{L^\frac{qp}{s(p-q)}_\omega}
    \lesssim\|G_{\mu,s}^v\|^s_{A^p_\omega\to L^q_\omega}.
    $$
This finishes the proof.
\end{proof}







\section{Compact operators $G_{\mu,s}^v:A^p_\omega\to L^q_\omega$}

In this section we prove Theorem~\ref{Theorem:main-s}(ii).

\begin{lemma}\label{technichal}
Let $\nu$ be a finite positive  Borel measure on $\mathbb{D}$ and $0<p<\infty$. If $\{\vp_n\}_{n=0}^\infty\subset L^p_\nu$ and $\vp\in  L^p_\nu$ satisfy
 $\lim_{n\to\infty}\|\vp_n\|_{L^p_\nu}=\|\vp\|_{L^p_\nu}$ and
$\lim_{n\to\infty}\vp_n(z)=\vp(z)$ $\nu$-a.e. on $\mathbb{D}$,
then $\lim_{n\to\infty}\|\vp_n-\vp\|_{L^p_\nu}=0$.
\end{lemma}

\begin{proof}
See the proof of \cite[Lemma~1~p.~21]{DurenHp}.
\end{proof}

\begin{theorem}\label{th:c1}
Let $0<p\le q<\infty$, $0<s<\infty$ and $\omega\in\widehat{\mathcal{D}}$, and let $\mu,v$ be  positive Borel measures on $\mathbb{D}$
such that $\mu\left(\{z\in\mathbb{D}: v(T(z))=0\}\right)=0=\mu(\{0\})$. Then $G_{\mu,s}^v : A^p_\omega \to L^q_\omega$ is compact if and only if
$I_d:\,A^p_\omega\to L^{p+s-\frac{ps}{q}}_{\mu^{\om}_v}$ is compact.
\end{theorem}

\begin{proof}
Let first $q>s$. Assume that $G_{\mu,s}^v:A^p_\omega\to L^q_\omega$ is compact.
For each $a\in\mathbb{D}$, let
    $
    f_{a,p}(z)
    =\left(\om(S(a))\right)^{-\frac1p}F_{a,p}(z)
    =\left(\om(S(a))\right)^{-\frac1p}\left(\frac{1-|a|^2}{1-\overline{a}z}\right)^{\frac{\g+1}{p}}
    $.
By Lemma~\ref{le:1} we may choose $\gamma=\gamma(p,q,s,\om)$ sufficiently large such that $\sup_{a\in\mathbb{D}}\|f_{a,p}\|_{A^p_\omega}^p\asymp1$,
$\sup_{a\in\mathbb{D}}\|f_{a,(\frac qs)'}\|_{A^{(\frac qs)'}_\omega}^{(\frac qs)'}\asymp1$ and $f_{a,p}$ converges uniformly to zero on compact subsets of $\mathbb{D}$ as $|a|\to 1^-$.
A standard argument shows that
    \begin{equation}\label{eq:c1}
    \lim_{|a|\to 1^-}\|G_{\mu,s}^v (f_{a,p})\|_{L^{q}_\omega}=0.
    \end{equation}
Fubini's theorem and H\"older's inequality yield
    \begin{equation*}
    \begin{split}
    \frac{\mu^{\om}_v(S(a))}{\omega(S(a))^{1+\frac{s}{p}-\frac{s}{q}}}&\asymp
    \frac{1}{\omega(S(a))^{1-\frac{s}{q}}}\int_{S(a)} |f_{a,p}(z)|^s\,d\mu^{\om}_v(z)\\
    &\asymp\int_{S(a)}|f_{a,p}(z)|^s\left(\frac{1}{v(T(z))}\int_{T(z)}|f_{a,(\frac qs)'}(\z)|\omega(\z)\,dA(\z)\right)d\mu(z)\\
    &\lesssim\int_\mathbb{D}|f_{a,(\frac qs)'}(\z)|\left(\int_{\Gamma(\z)}|f_{a,p}(z)|^s\frac{d\mu(z)}{v(T(z))}\right)\omega(\z)\,dA(\z)\\
    &\le\|f_{a,(\frac qs)'}\|_{A^{(\frac qs)'}_\omega}\|G_{\mu,s}^v (f_{a,p})\|^s_{L^{q}_\omega}
    \lesssim \|G_{\mu,s}^v (f_{a,p})\|^s_{L^{q}_\omega}.
    \end{split}
    \end{equation*}
This together with \eqref{eq:c1} gives
    $
    \lim_{|a|\to 1^-}\frac{\mu^{\om}_v(S(a))}{\omega(S(a))^{1+\frac{s}{p}-\frac{s}{q}}}=0
    $
and hence $I_d:A^p_\omega\to L^{p+s-\frac{ps}{q}}_{\mu^{\om}_v}$ is compact by Theorem~\ref{Theorem:normscm}.

Conversely, let $q>s$ and assume that $I_d:\,A^p_\omega\to L^{p+s-\frac{ps}{q}}_{\mu^{\om}_v}$  is compact.
Write $t=t(p,q,s)=p+s-\frac{ps}{q}>s$ for short.
Let $\{f_n\}$ be a bounded sequence in $A^p_\om$. Then, we may extract a subsequence
$\{f_{n_k}\}$ that converges in $L^t_{\mu^{\om}_v}$ and uniformly on compact subsets to some $f\in A^p_\om$.
Write $g_{n_k}=f_{n_k}-f$. By Fubini's theorem, \cite[Theorem~3]{JAJRTent} and H\"older's inequality,
    \begin{equation*}
    \begin{split}
    \|G_{\mu,s}^v(g_{n_k})\|^s_{L^{q}_\omega}&=\sup_{\|h\|_{L^{(\frac{q}{s})'}_\omega}\le1}\int_\mathbb{D}|h(z)|
    \left(\int_{\Gamma(z)}|g_{n_k}(\z)|^s\frac{d\mu^{\om}_v (\z)}{\omega(T(\z))}\right)\omega(z)\,dA(z)
    \lesssim \|g_{n_k}\|^s_{L^{t}_{\mu^{\om}_v}}
    \end{split}
    \end{equation*}
and hence $$
\lim_{k\to\infty}\|G_{\mu,s}^v(g_{n_k})\|^s_{L^{q}_\omega}=0.
$$
If $s\ge 1$, two applications of Minkowski's inequality gives
    $$
    \left|\|G_{\mu,s}^v(f_{n_k})\|_{L^{q}_\omega}-\|G_{\mu,s}^v(f)\|_{L^{q}_\omega} \right|\le \|G_{\mu,s}^v(g_{n_k})\|_{L^{q}_\omega}\to 0,\quad k\to \infty.
    $$
Moreover, since $\{f_{n_k}\}$  converges uniformly on compact subsets of $\mathbb{D}$ to $f$, then $\vp_k(z)=\left(\int_{\Gamma(z)}|f_{n_k}(\z)|^s\frac{d\mu^{\om}_v (\z)}{\omega(T(\z))}\right)^{1/s}$ converges to $\vp(z)=\left(\int_{\Gamma(z)}|f(\z)|^s\frac{d\mu^{\om}_v (\z)}{\omega(T(\z))}\right)^{1/s}$ for
each $z\in\mathbb{D}$. Therefore Lemma~\ref{technichal} yields
    $$
    \lim_{k\to\infty}\|G_{\mu,s}^v(f_{n_k})-G_{\mu,s}^v(f)\|_{L^{q}_\omega}
    =\lim_{k\to\infty}\|\vp_k-\vp\|_{L^{q}_\omega}=0,
    $$
and thus $G_{\mu,s}^v:A^p_\omega\to L^q_\omega$ is compact.

If $0<s\le 1$,
    $$
    \int_{\Gamma(z)}|f_{n_k}(\z)|^s\frac{d\mu^{\om}_v (\z)}{\omega(T(\z))}\le \int_{\Gamma(z)}|g_{n_k}(\z)|^s\frac{d\mu^{\om}_v (\z)}{\omega(T(\z))}+\int_{\Gamma(z)}|f(\z)|^s\frac{d\mu^{\om}_v (\z)}{\omega(T(\z))},
    $$
which together with Minkowski's inequality yields
    $$
    \left|\|G_{\mu,s}^v(f_{n_k})\|^s_{L^{q}_\omega}-\|G_{\mu,s}^v(f)\|^s_{L^{q}_\omega} \right|\le \|G_{\mu,s}^v(g_{n_k})\|^s_{L^{q}_\omega}.
    $$
Now, by arguing as in the previous case we see that $G_{\mu,s}^v:A^p_\omega\to L^q_\omega$ is compact. In view of~\eqref{Eq:case-q=s}, a similar reasoning also applies in the case $q=s$.

It remains to consider the case $q<s$. Assume first that $I_d:A^p_\omega\to L^{p+s-\frac{ps}{q}}_{\mu^{\om}_v}$ is compact.
Let $\{f_n\}$ be a bounded sequence in $A^p_\om$. Then we may extract a subsequence
$\{f_{n_k}\}$ that converges on $L^{p+s-\frac{ps}{q}}_{\mu^{\om}_v}$ and uniformly on compact subsets of $\mathbb{D}$ to
some $f\in A^p_\om$. Write $g_{n_k}=f_{n_k}-f$, and let $0<x<s$. H\"older's inequality and Fubini's theorem yield
    \begin{equation*}
    \begin{split}
    \|G_{\mu,s}^v(g_{n_k})\|^q_{L^{q}_\omega}
    &\le\int_\mathbb{D} N(g_{n_k})(z)^{\frac{qx}{s}}\left(\int_{\Gamma(z)}|g_{n_k}(\z)|^{s-x}\frac{d\mu^{\om}_v(\z)}{\omega(T(\z))}\right)^\frac{q}{s}\omega(z)\,dA(z)\\
    & \lesssim\|N(g_{n_k})\|^{\frac{qx}{s}}_{L_\omega^{\frac{qx}{s-q}}}\|g_{n_k}\|^{\frac{q(s-x)}{s}}_{L_{\mu^{\om}_v}^{(s-x)}}.
    \end{split}
    \end{equation*}
Take $x=\frac{p(s-q)}{q}<s$ so that $s-x=s+p-\frac{ps}{q}$. Then the estimates above together with \cite[Lemma 4.4]{PelRat}  give
    \begin{equation*}
    \begin{split}
    \|G_{\mu,s}^v(g_{n_k})\|^q_{L^{q}_\omega}
    \lesssim\|g_{n_k}\|^{\frac{qx}{s}}_{A_\omega^{p}}\|g_{n_k}\|^{\frac{q(s-x)}{s}}_{L_{\mu^{\om}_v}^{s+p-\frac{ps}{q}}}
    \lesssim \|g_{n_k}\|^{\frac{q(s-x)}{s}}_{L_{\mu^{\om}_v}^{s+p-\frac{ps}{q}}},
    \end{split}
    \end{equation*}
and hence
    $$
    \lim_{k\to\infty}\|G_{\mu,s}^v(g_{n_k})\|_{L^{q}_\omega}=0.
    $$
Now, by using Lemma~\ref{technichal} and arguing as in the case $q>s$, we conclude that
$\lim_{k\to\infty}\|G_{\mu,s}^v(f_{n_k})-G_{\mu,s}^v(f)\|_{L^{q}_\omega}
= 0$, that is, $G_{\mu,s}^v:A^p_\omega\to L^q_\omega$ is compact.

Conversely, let $q<s$ and assume that $G_{\mu,s}^v:A^p_\omega\to L^q_\omega$ is bounded. Choose $\alpha>\beta>1$ such that $\frac{\beta}{\alpha}=\frac{q}{s}$. By arguing as in \eqref{1-s} and \eqref{1-s'}, we get
    \begin{equation*}
    \begin{split}
    \frac{\mu^{\om}_v(S(a))}{\om\left(S(a)\right)^{s/p}}
    &\lesssim\|G_{\mu,s}^v(f_{a,p})\|^{\frac{q}{\beta}}_{L^{q}_\omega}
    \|G_{\mu,1}^v(f^s_{a,p}(z)\chi_{S(a)})\|^{\frac{1}{\alpha'}}_{L^{\frac{\beta'}{\alpha'}}_\omega}
    \end{split}
    \end{equation*}
and
    \begin{equation*}
    \begin{split}
    \|G_{\mu,1}^v(f^s_{a,p}(z)\chi_{S(a)})\|_{L^{\frac{\beta'}{\alpha'}}_\omega}
    &\lesssim\frac{\mu^{\om}_v(S(a))^{\frac{\alpha'}{\beta'}}}{\om\left(S(a)\right)^{s/p}} \left(\sup_{z\in \mathbb{D}} \frac{\mu^{\om}_v(S(z)\cap S(a))}{\omega(S(z))} \right)^{1-\frac{\alpha'}{\beta'}},\quad a\in\mathbb{D},
    \end{split}
    \end{equation*}
respectively. These estimates yield
    \begin{equation*}
    \begin{split}
    \frac{\mu^{\om}_v(S(a))}{\om\left(S(a)\right)^{s/p}}
    \lesssim \|G_{\mu,s}^v(f_{a,p})\|^{\frac{q}{\beta}}_{L^{q}_\omega}\frac{\mu^{\om}_v(S(a))^{\frac{1}{\beta'}}}{\om\left(S(a)\right)^{\frac{s}{p\alpha'}}} \left(\sup_{z\in \mathbb{D}} \frac{\mu^{\om}_v(S(z)\cap S(a))}{\omega(S(z))} \right)^{\frac{1}{\alpha'}-\frac{1}{\beta'}},\quad a\in\mathbb{D},
    \end{split}
    \end{equation*}
and thus
    \begin{equation}
    \begin{split}\label{eq:1-sc}
    \frac{\mu^{\om}_v (S(a))}{\omega(S(a))^{\frac{q}{p}}}
    &\lesssim\|G_{\mu,s}^v(f_{a,p})\|^{q}_{L^{q}_\omega}
    \left(\sup_{z\in \mathbb{D}} \frac{\mu^{\om}_v(S(z)\cap S(a))}{\omega(S(z))} \right)^{\frac{\beta}{\alpha'}-\frac{\beta}{\beta'}},
    \quad a\in\mathbb{D}.
    \end{split}
    \end{equation}
If $q=p$, we may use $\lim_{|a|\to 1^-}\|G_{\mu,s}^v (f_{a,p})\|_{L^{p}_\omega}=0$ and Theorem~\ref{th:casepleq} to deduce that the right-hand side of \eqref{eq:1-sc} tends to zero as $a$ approaches the boundary. Therefore $I_d:A^p_\om\to L^p_{\mu^{\om}_v}$ is compact by Theorem~\ref{Theorem:normscm}(ii).

If $q>p$, by using \eqref{eq:1-sc} and arguing as in the corresponding part of the proof of Theorem~\ref{th:casepleq},
we get
    \begin{equation*}
    \begin{split}
    \frac{\mu^{\om}_v(S(a))}{\omega(S(a))^{1+\frac{s}{p}-\frac{s}{q}}}
    &\le\|G_{\mu,s}^v(f_{a,p})\|^{q}_{L^{q}_\omega}
    \left(\sup_{b\in\mathbb{D}}\frac{\mu^{\om}_v(S(b))}{\omega(S(b))^{1+\frac{s}{p}-\frac{s}{q}}}\right)^{1-\frac{q}{s}},
    \quad a\in\mathbb{D}.
    \end{split}
    \end{equation*}
from which arguments similar to those applied in the previous paragraph show that $I_d:A^p_\om\to L^{p+s-\frac{ps}{q}}_{\mu^{\om}_v}$ is compact.
This finishes the proof.
\end{proof}

\begin{theorem}
Let $0<q<p<\infty$ and $0<s<\infty$ such that $1+\frac{s}p-\frac{s}{q}>0$, $\omega\in\widehat{\mathcal{D}}$ and let $\mu,v$ be  positive Borel measures on $\mathbb{D}$
such that $\mu\left(\{z\in\mathbb{D}: v(T(z))=0\}\right)=0=\mu(\{0\})$. Then the following conditions are equivalent:
    \begin{enumerate}
    \item[\rm(i)] $G_{\mu,s}^v:A^p_\omega\to L^q_\omega$ is compact;
    \item[\rm(ii)] $G_{\mu,s}^v:A^p_\omega\to L^q_\omega$ is bounded;
    \item[\rm(iii)] $I_d: A^p_\om\to L^{p\left(1+\frac{s}p-\frac{s}{q}\right)}_{\mu^{\om}_v}$ is compact;
    \item[\rm(iv)] $I_d: A^p_\om\to L^{p\left(1+\frac{s}p-\frac{s}{q}\right)}_{\mu^{\om}_v}$ is bounded.
    \end{enumerate}
\end{theorem}

\begin{proof}
The conditions (ii)--(iv) are equivalent by Theorems~\ref{Theorem:main-s}(i) and~\ref{Theorem:normscm}(iii). To complete the proof, it suffices to show that $G_{\mu,s}^v:A^p_\omega\to L^q_\omega$ is compact if $I_d: A^p_\om\to L^{p\left(1+\frac{s}p-\frac{s}{q}\right)}_{\mu^{\om}_v}$ is bounded. To see this, let $\{f_n\}$ be a bounded sequence in $A^p_\om$, let $\{f_{n_k}\}$ be a subsequence that converges uniformly on compact subsets of $\mathbb{D}$ to $f\in A^p_\om$. Write $g_k=f_{n_k}-f$ as before. By using Theorem~\ref{Theorem:normscm} and the last part of the proof of Theorem~\ref{Theorem:normscm}(iii), we deduce
    $$
    \lim_{R\to 1^-} \int_\mathbb{D} \left(\int_{\Gamma(\z)\setminus\overline{D(0,R)}} \frac{d\mu(z)}{\om(T(z))}\right)^{\frac{pq}{s(p-q)}}\,\om(\z)\,dA(\z)=0.
    $$
Therefore, for a fixed $\e>0$, there exists $R_0\in(0,1)$ such that
    $$
    \int_\mathbb{D} \left(\int_{\Gamma(\z)\setminus\overline{D(0,R_0)}} \frac{d\mu(z)}{\om(T(z))}\right)^{\frac{pq}{s(p-q)}}\,\om(\z)\,dA(\z)<\e^\frac{p}{p-q}.
    $$
Choose $k_0\in\mathbb{N}$ such that $|g_k(z)|<\e^{1/q}$ for all $k\ge k_0$ and $z\in\overline{D(0,R_0)}$. Then H\"older's inequality and \cite[Lemma~4.4]{PelRat} give
    \begin{equation*}
    \begin{split}
    \|G_{\mu,s}^v(g_k)\|^q_{L^{q}_\omega}
    &\lesssim  \int_\mathbb{D}\left(\int_{\Gamma(z)\cap\overline{D(0,R_0)}}|g_k(\z)|^s\frac{d\mu(\z)}{v(T(\z))}\right)^\frac{q}{s}\omega(z)\,dA(z)\\
    &\quad+\int_\mathbb{D}\left(\int_{\Gamma(z)\setminus\overline{D(0,R_0)}}|g_k(\z)|^s\frac{d\mu(\z)}{v(T(\z))}\right)^\frac{q}{s}\omega(z)\,dA(z)\\
    &\lesssim \e+
    \int_\mathbb{D} N(g_k)^q(z)\left(\int_{\Gamma(z)\setminus\overline{D(0,R_0)}}\frac{d\mu(\z)}{v(T(\z))}\right)^\frac{q}{s}\omega(z)\,dA(z)\\
    &\le \e+
    \|N(g_k) \|^q_{L^{p}_\omega}\left(\int_\mathbb{D} \left(\int_{\Gamma(z)\setminus\overline{D(0,R_0)}}\frac{d\mu(\z)}{v(T(\z))}\right)^\frac{pq}{(p-q)s}\omega(z)\,dA(z)\right)^\frac{p-q}{p}
    \lesssim \e,
    \end{split}
    \end{equation*}
and consequently, $\lim_{k\to \infty}\|G_{\mu,s}^v(g_k)\|^q_{L^{q}_\omega}=0$. Finally, by using Lemma~\ref{technichal} and arguing as in the proof of Theorem~\ref{th:c1}, we deduce $\lim_{k\to \infty}\|G_{\mu,s}^v(f_k)-G_{\mu,s}^v(f)\|^q_{L^{q}_\omega}=0$, that is, $G_{\mu,s}^v:A^p_\omega\to L^q_\omega$ is compact.
\end{proof}






\section{Applications and further comments}\label{Sec:comments}

\subsection{Area operators in Hardy spaces}\label{Subsection:Area-Hardy}

For $0<s<\infty$, define
    $$
    G_{\mu,s}(f)(z)=\left(\int_{\Gamma(z)}|f(\z)|^s\frac{d\mu (\z)}{1-|\z|}\right)^\frac1s,\quad z\in\mathbb{T}.
    $$
The method of proof of Theorem~\ref{Theorem:main-s}, combined with the results in \cite[Section $7$]{Lu90}
and \cite{PelPAMS14}, can be used to obtain the following result. The details of the proof does not reveal anything new, and are therefore omitted. Here $L^q(\mathbb{T})$ refers to the classical $L^q$-space on $\mathbb{T}$.

\begin{theorem}\label{Theorem:main-sHp}
Let $0<p,q<\infty$ such that $1+\frac{s}p-\frac{s}{q}>0$, and let $\mu$ be a positive Borel measure on $\mathbb{D}$ such that $\mu(\{0\})=0$. Then $G_{\mu,s}:H^p\to L^q(\mathbb{T})$ is bounded
 (resp. compact) if and only if $I_d:\,H^p\to L^{p+s-\frac{ps}{q}}_\mu$ is bounded (resp. compact).
Moreover,
    \begin{equation*}
    \begin{split}
    \|G_{\mu,s}\|^s_{H^p\to L^{q}(\mathbb{T})}
    \asymp\|Id\|^{p+s-\frac{ps}{q}}_{H^p\to L^{p+s-\frac{ps}{q}}_{\mu}}
    \asymp\sup_{a\in\mathbb{D}}\frac{\mu(S(a))}{(1-|a|)^{1+\frac{s}{p}-\frac{s}{q}}},\quad p\le q,
    \end{split}
    \end{equation*}
and
    \begin{equation*}
    \begin{split}
    \|G_{\mu,s}\|^s_{H^p\to L^{q}(\mathbb{T})}
    \asymp\|Id\|^{p+s-\frac{ps}{q}}_{H^p\to L^{p+s-\frac{ps}{q}}_{\mu}}
    \asymp\|B_\mu\|_{L^{\frac{qp}{s(p-q)}}},\quad q<p,
    \end{split}
    \end{equation*}
where
    $$
    B_\mu(\z)=\int_{\Gamma(\z)}\frac{d\mu(z)}{1-|z|},\quad
    \z\in\mathbb{T}.
    $$
\end{theorem}

In particular, this result proves the conjecture in \cite[p.~365]{GonLouWu2010} in the case $1+\frac{1}p-\frac{1}{q}>0$.





\subsection{Integral operator $T_g$ on Bergman and Hardy spaces}

Each $g\in\H(\mathbb{D})$ induces the integral operator
    $$
    T_g(f)(z)=\int_0^zg'(\z)f(\z)\,d\z,\quad z\in\mathbb{D},
    $$
acting on $\H(\mathbb{D})$. This type of integral operators have been extensively studied
during the last decades and have interesting connections with other areas of mathematical analysis, see \cite{PelRat,PelSum14} and the references therein. In particular, the symbols $g$ for which $T_g$ is bounded or compact from $A^p_\om$ to
$A^q_\om$ can be described in terms of the following spaces of analytic functions when $q\ge p$.

We say that $g\in\H(\mathbb{D})$ belongs to $\mathcal{C}^{q,p}(\om^\star)$,
$0<p,q<\infty$, if the measure $|g'(z)|^2\om^\star(z)\,dA(z)$ is a
$q$-Carleson measure for $A^p_\om$. Moreover,
$g\in\mathcal{C}^{q,p}_0(\om^\star)$ if the identity operator
$I_d:A^p_\omega\to L^q(|g'|^2\omega^\star dA)$ is compact. If
$q\ge p$ and $\om\in\widehat{\mathcal{D}}$, then Theorem~\ref{Theorem:normscm} shows that
these spaces only depend on the quotient $\frac{q}{p}$.
Consequently, for $q\ge p$ and $\om\in\widehat{\mathcal{D}}$, we simply write
$\mathcal{C}^{q/p}(\om^\star)$ instead of $\mathcal{C}^{q,p}(\om^\star)$. Thus,
if $\alpha\ge 1$ and $\om\in\widehat{\mathcal{D}}$, then
$\mathcal{C}^{\alpha}(\om^\star)$ consists of those $g\in\H(\mathbb{D})$ such
that
    \begin{equation}\label{calpha}
    \|g\|^2_{\mathcal{C}^{\alpha}(\om^\star)}=|g(0)|^2+\sup_{I\subset\mathbb{T}}\frac{\int_{S(I)}|g'(z)|^2\om^\star(z)\,dA(z)}
    {\left(\om\left(S(I)\right)\right)^{\alpha}}<\infty.\index{$\mathcal{C}^\alpha}       \def\b{\beta}        \def\g{\gamma(\om^\star)$}
    \end{equation}
An analogue of this identity is valid for the little space
$\mathcal{C}^{\alpha}_0(\om^\star)$. We refer to \cite[Chapter~5]{PelRat}
for further information about these spaces.



\begin{theorem}\label{Theorem:T_g-Bergman}
Let $0<p,q<\infty$ such that $q>\frac{2p}{2+p}$ and $\om\in\widehat{\mathcal{D}}$. Let $g\in \H(\mathbb{D})$ and denote $d\mu_g(z)=|g'(z)|^2\om^\star(z)\,dA(z)$. Then $T_g:A^p_\om\to A^q_\om$ is bounded (resp. compact) if and only if $I_d:A^p_\omega\to L^{p+2-\frac{2p}{q}}_{\mu_g}$ is bounded (resp. compact).
\end{theorem}

\begin{proof}
By \cite[Theorem~4.2]{PelRat}, $T_g:A^p_\om\to L^q_\om$ is bounded  (resp. compact) if and only if $G_{\mu_g,2}^\om:A^p_\om\to L^q_\om$ is bounded (resp. compact), and
    $
    \|T_g\|_{A^p_\om\to L^q_\om}\asymp\|G_{\mu_g,2}^\om\|_{A^p_\om\to L^q_\om}
    $.
Theorem~\ref{Theorem:main-s}
implies
    $$
    \|G_{\mu_g,2}^\om\|^2_{A^p_\om\to L^q_\om}
    \asymp\|I_d\|^{p+2-\frac{2p}{q}}_{A^p_\omega\to L^{p+2-\frac{2p}{q}}_{\mu_g}},
    $$
and this finishes the proof.
\end{proof}

It is worth mentioning that Theorem~\ref{Theorem:normscm} yields
    $$
    \|T_g\|_{A^p_\om\to L^q_\om}^2
    \asymp\sup_{a\in\mathbb{D}}\frac{\mu_g(S(a))}{\om(S(a))^{2(\frac1p-\frac1q)+1}},\quad q\ge p
    $$
Thus $T_g:A^p_\om\to L^q_\om$ is bounded if and only if $g\in\mathcal{C}^{2(\frac1p-\frac1q)+1}(\om^\star)$. If $p>q$, Theorem~\ref{Theorem:normscm} also gives
    $$
    \|T_g\|_{A^p_\om\to L^q_\om}^2
    \asymp\int_\mathbb{D}\left(\int_{\Gamma(\z)}|g'(z)|^2\,dA(z)\right)^\frac{qp}{2(p-q)}\om(\z)\,dA(\z),
    $$
and thus $T_g:A^p_\om\to L^q_\om$ is bounded if and only if $g\in A^\frac{qp}{p-q}_\om$ by \cite[Theorem~4.2]{PelRat}.

Consequently, whenever $q>\frac{2p}{2+p}$
Theorem~\ref{Theorem:T_g-Bergman} improves \cite[Theorem~4.1]{PelRat} because the hypothesis on $\om$ are stronger in the original result.
In particular, if $q<p$, the weight $\om$ is assumed to be continuous and strictly positive with the local regularity
    $$
    \om(t)\asymp\om(r),\quad 1-t\asymp1-r.
    $$
This hypothesis allows one to use the strong factorization $A^p_\om=A^{p_1}_\om\cdot A^{p_2}_\om$, $p^{-1}=p_1^{-1}+p_2^{-1}$,
\cite[Theorem~3.1]{PelRat}, which is a principal ingredient in the proof of \cite[Theorem~4.1]{PelRat}.

However, the defect of Theorem~\ref{Theorem:T_g-Bergman} is the extra hypothesis $q>\frac{2p}{2+p}$ which is a restriction only in the case $p>q$. This condition is inherited from Theorem~\ref{Theorem:main-s} and appears there because Carleson measures are finite measures. This is not true in general for~$\mu_g$ when $g\in A^\frac{qp}{p-q}_\om$ and $q<\frac{2p}{2+p}$.
The case of compact operators can be analyzed in the same way.

If $q>\frac{2p}{2+p}$  one may characterize bounded and compact operators $T_g:H^p\to H^q$ by using Section~\ref{Subsection:Area-Hardy}. In order to avoid unnecessary repetition, we omit the details.




