Compact differences of weighted composition operators

Abstract

Compact differences of two weighted composition operators acting from the weighted Bergman space $A^p_{\omega }$ to another weighted Bergman space $A^q_{\nu }$, where $0<p\le q<\infty$ and $\omega ,\nu$ belong to the class ${\mathcal {D}}$ of radial weights satisfying two-sided doubling conditions, are characterized. On the way to the proof a new description of q-Carleson measures for $A^p_{\omega }$, with $\omega \in {\mathcal {D}}$, in terms of pseudohyperbolic discs is established. This last-mentioned result generalizes the well-known characterization of q-Carleson measures for the classical weighted Bergman space $A^p_{\alpha }$ with $-1<\alpha <\infty$ to the setting of doubling weights.

1 Introduction and main results

Let ${\mathcal {H}}({\mathbb {D}})$ denote the space of analytic functions in the unit disc ${\mathbb {D}}=\{z\in {\mathbb {C}}:|z|<1\}$. For a nonnegative function $\omega \in L^1([0,1))$, the extension to ${\mathbb {D}}$, defined by $\omega (z)=\omega (|z|)$ for all $z\in {\mathbb {D}}$, is called a radial weight. For $0<p<\infty$ and a radial weight $\omega$, the weighted Bergman space $A^p_{\omega }$ consists of $f\in {\mathcal {H}}({\mathbb {D}})$ such that

$$\begin{aligned} \Vert f\Vert _{A^p_{\omega }}^p=\int _{{\mathbb {D}}}|f(z)|^p\omega (z)\,dA(z)<\infty , \end{aligned}$$

where $dA(z)=\frac{dx\,dy}{\pi }$ is the normalized Lebesgue area measure on ${\mathbb {D}}$. As usual, $A^p_{\alpha }$ stands for the classical weighted Bergman space induced by the standard radial weight $\omega (z)=(1-|z|^2)^{\alpha }$, where $-1<\alpha <\infty$.

For a radial weight $\omega$, write $\widehat{\omega }(z)=\int _{|z|}^1\omega (s)\,ds$ for all $z\in {\mathbb {D}}$. In this paper we always assume $\widehat{\omega }(z)>0$, for otherwise $A^p_{\omega }={\mathcal {H}}({\mathbb {D}})$ for each $0<p<\infty$. A weight $\omega$ belongs to the class $\widehat{{\mathcal {D}}}$ if there exists a constant $C=C(\omega )\ge 1$ such that $\widehat{\omega }(r)\le C\widehat{\omega }(\frac{1+r}{2})$ for all $0\le r<1$. Moreover, if there exist $K=K(\omega )>1$ and $C=C(\omega )>1$ such that $\widehat{\omega }(r)\ge C\widehat{\omega }\left( 1-\frac{1-r}{K}\right)$ for all $0\le r<1$, then we write $\omega \in \check{{\mathcal {D}}}$. In other words, $\omega \in \check{{\mathcal {D}}}$ if there exists $K=K(\omega )>1$ and $C'=C'(\omega )>0$ such that

$$\begin{aligned} \widehat{\omega }(r)\le C'\int _r^{1-\frac{1-r}{K}}\omega (t)\,dt,\quad 0\le r<1. \end{aligned}$$

The intersection $\widehat{{\mathcal {D}}}\cap \check{{\mathcal {D}}}$ is denoted by ${\mathcal {D}}$, and this is the class of weights that we mainly work with.

Each analytic self-map $\varphi$ of ${\mathbb {D}}$ induces the composition operator $C_{\varphi }$ on ${\mathcal {H}}({\mathbb {D}})$ defined by $C_{\varphi } f=f\circ \varphi$. The weighted composition operator induced by $u\in {\mathcal {H}}({\mathbb {D}})$ and $\varphi$ is $u C_{\varphi }$ and sends $f\in {\mathcal {H}}({\mathbb {D}})$ to $u\cdot f\circ \varphi \in {\mathcal {H}}({\mathbb {D}})$. These operators have been extensively studied in a variety of function spaces (see for example [4,5,6,7,8, 22, 23, 25, 26]).

If now $\psi$ is another analytic self-map of ${\mathbb {D}}$, the pair $(\varphi ,\psi )$ induces the operator $C_{\varphi }-C_{\psi }$. One of the most important problem considering these operators is to characterize compact differences in Hardy spaces. Shapiro and Sundberg [24] studied this problem in 1990. Very recently, Choe, Choi, Koo and Yang [3] have solved this problem. For more about difference operators, see [2, 8, 10, 11, 20]. Moorhouse [11, 12] obtain some important results on this operator in weighted Bergman spaces. He showed [11], among other things, that $C_{\varphi }-C_{\psi }$ is compact on $A^2_{\alpha }$ if and only if

$$\begin{aligned} \lim _{|z|\rightarrow 1^-}|\delta _1(z)|\left( \frac{1-|z|^2}{1-|\varphi (z)|^2} +\frac{1-|z|^2}{1-|\psi (z)|^2}\right) = 0, \end{aligned}$$

(1)

where

$$\begin{aligned} \delta _1(z)=\frac{\varphi (z)-\psi (z)}{1-\overline{\varphi (z)}\psi (z)},\quad z\in {\mathbb {D}}. \end{aligned}$$

Saukko [20, 21] generalized this result by showing that if either $1<p\le q$, or $p>q\ge 1$, then $C_{\varphi }-C_{\psi }:A^p_{\alpha }\rightarrow A^q_{\beta }$ is compact if and only if the operators $\delta _1C_{\varphi }$ and $\delta _1C_{\psi }$ are both compact from $A^p_{\alpha }$ to $L^q_{\beta }$. Very recently, Acharyya and Wu [1] characterized the compact differences of two weighted composition operators $uC_{\varphi }-vC_{\psi }$ between different weighted Bergman spaces $A^p_{\alpha }$ and $A^q_{\beta }$, where $0<p\le q< \infty$ and $-1<\alpha ,\beta <\infty$. Their result states that, if $\frac{\alpha +2}{p}\le \frac{\beta +2}{q}$ and u, $v\ \in {\mathcal {H}}({\mathbb {D}})$ satisfy

$$\begin{aligned} \sup _{z\in {\mathbb {D}}}\left( |u(z)|+|v(z)|\right) (1-|z|^2)^{\frac{\beta +2}{q} -\frac{\alpha +2}{p}}<\infty , \end{aligned}$$

then $uC_{\varphi }-vC_{\psi }:A^p_{\alpha }\rightarrow A^q_{\beta }$ is compact if and only if

$$\begin{aligned} \lim _{|z|\rightarrow 1^-}|\delta _1(z)|\left( |u(z)| \frac{(1-|z|^2)^{\frac{2+\beta }{q}}}{(1-|\varphi (z)|^2)^{\frac{2+\alpha }{p}}} + |v(z)|\frac{(1-|z|^2)^{\frac{2+\beta }{q}}}{(1-|\psi (z)|^2)^{\frac{2+\alpha }{p}}}\right) = 0 \end{aligned}$$

(2)

and

$$\begin{aligned} \lim _{|z|\rightarrow 1^-}(1-|\delta _1(z)|^2)^{\frac{2+\alpha }{p}}|u(z) -v(z)|\left( \frac{(1-|z|^2)^{\frac{2+\beta }{q}}}{(1-|\varphi (z)|^2)^{\frac{2+\alpha }{p}}} +\frac{(1-|z|^2)^{\frac{2+\beta }{q}}}{(1-|\psi (z)|^2)^{\frac{2+\alpha }{p}}}\right) = 0. \end{aligned}$$

(3)

In this paper we characterize compact differences of two weighted composition operators from the weighted Bergman space $A^p_{\omega }$ to another weighted Bergman space $A^q_{\nu }$ with $0<p\le q<\infty$ and $\omega ,\nu \in {\mathcal {D}}$. To state the result, write

$$\begin{aligned} \delta _2(z)=\frac{\psi (z)-\varphi (z)}{1-\overline{\psi (z)}\varphi (z)},\quad z\in {\mathbb {D}}, \end{aligned}$$

and observe that $|\delta _1|=|\delta _2|$ on ${\mathbb {D}}$. Our main result reads as follows.

Theorem 1

Let $\omega ,\nu \in {\mathcal {D}}$ and $0<p\le q<\infty$ such that $\widehat{\nu }(z)(1-|z|)\lesssim \left( \widehat{\omega }(z)(1-|z|)\right) ^{\frac{q}{p}}$ for all $z\in {\mathbb {D}}$. Further, let $u,v\in {\mathcal {H}}({\mathbb {D}})$ and $\varphi$ and $\psi$ analytic self-maps of ${\mathbb {D}}$ such that

$$\begin{aligned} \sup _{z\in {\mathbb {D}}}\left( |u(z)|+|v(z)|\right) \frac{\left( \widehat{\nu }(z) (1-|z|)\right) ^{\frac{1}{q}}}{\left( \widehat{\omega }(z)(1-|z|)\right) ^{\frac{1}{p}}}<\infty . \end{aligned}$$

(4)

Then there exists $\gamma =\gamma (\omega ,p)>0$ with the following property: $uC_{\varphi }-vC_{\psi }:A^p_{\omega }\rightarrow A^q_{\nu }$ is compact if and only if

$$\begin{aligned} \lim _{|z|\rightarrow 1^-}|\delta _1(z)| \left( |u(z)|\frac{\left( \widehat{\nu }(z)(1-|z|)\right) ^{\frac{1}{q}}}{\left( \widehat{\omega }(\varphi (z))(1-|\varphi (z)|)\right) ^{\frac{1}{p}}} +|v(z)|\frac{\left( \widehat{\nu }(z)(1-|z|)\right) ^{\frac{1}{q}}}{\left( \widehat{\omega }(\psi (z))(1-|\psi (z)|)\right) ^{\frac{1}{p}}}\right) = 0 \end{aligned}$$

(5)

and

$$\begin{aligned}&\lim _{|z|\rightarrow 1^-}\left( \frac{|1-\overline{\varphi (z)}\delta _1(z)|^{\gamma }}{\left( \widehat{\omega }(\varphi (z))(1-| \varphi (z)|)\right) ^{\frac{1}{p}}} +\frac{|1-\overline{\psi (z)}\delta _2(z)|^{\gamma }}{\left( \widehat{\omega }(\psi (z))(1-| \psi (z)|)\right) ^{\frac{1}{p}}}\right) \nonumber \\&\quad \cdot |u(z)-v(z)|\left( \widehat{\nu }(z)(1-|z|)\right) ^{\frac{1}{q}}=0. \end{aligned}$$

(6)

If $\omega (z)=(1-|z|)^{\alpha }$ and $\nu (z)=(1-|z|)^{\beta }$ for $-1<\alpha ,\beta <\infty$, then $\widehat{\omega }(z)\asymp (1-|z|)^{\alpha +1}$ and $\widehat{\nu }(z)\asymp (1-|z|)^{\beta +1}$ for all $z\in {\mathbb {D}}$. Therefore (5) reduces to (2). Moreover, the Proof of Theorem 1 shows that the only requirement for $\gamma =\gamma (\omega ,p)>0$ appearing in the statement is that

$$\begin{aligned} \int _{{\mathbb {D}}}\frac{\omega (z)}{|1-\overline{a}z|^{\gamma p}}\,dA(z) \le C\frac{\widehat{\omega }(a)}{(1-|a|)^{\gamma p-1}},\quad a\in {\mathbb {D}}, \end{aligned}$$

for some constant $C=C(\omega ,p,\gamma )>0$. If $\omega (z)=(1-|z|)^{\alpha }$, any $\gamma >\frac{\alpha +2}{p}$ is acceptable, and the choice $\gamma =2\frac{\alpha +2}{p}$ converts (6) to (3), as a simple computation shows. Therefore Theorem 1 indeed generalizes [1, Theorem 1] for weights in ${\mathcal {D}}$.

We need two specific tools for the Proof of Theorem 1. The first one concerns continuous embeddings $A^p_{\omega }\subset L^q_{\mu }$. Recall that a positive Borel measure $\mu$ on ${\mathbb {D}}$ is a q-Carleson measure for $A^p_{\omega }$ if the identity operator $I_d:A^p_{\omega }\rightarrow L^q_{\mu }$ is bounded. A complete characterization of such measures in the case $\omega \in \widehat{{\mathcal {D}}}$ can be found in [15], see also [13, 17]. In particular, it is known that if $q\ge p$ and $\omega \in \widehat{{\mathcal {D}}}$, then $\mu$ is a q-Carleson measure for $A^p_{\omega }$ if and only if

$$\begin{aligned} \sup _{a\in {\mathbb {D}}}\frac{\mu (S(a))}{\omega (S(a))^{\frac{q}{p}}}<\infty . \end{aligned}$$

Here and from now on $S(a)=\{z:1-|a|<|z|<1,\,|\arg z-\arg a|<(1-|a|)/2\}$ is the Carleson square induced by the point $a\in {\mathbb {D}}\setminus \{0\}$, $S(0)={\mathbb {D}}$ and $\omega (E)=\int _E\omega dA$ for each measurable set $E\subset {\mathbb {D}}$. We will need a variant of this result and its “compact” counterpart for $\omega \in {\mathcal {D}}$ where the Carleson squares are replaced by pseudohyperbolic discs. To this end, denote $\varphi _a(z)=\frac{a-z}{1-\overline{a}z}$ for $a,z\in {\mathbb {D}}$. The pseudohyperbolic distance between two points a and b in ${\mathbb {D}}$ is $\rho (a,b)=|\varphi _a(b)|$. For $a\in {\mathbb {D}}$ and $0<r<1$, the pseudohyperbolic disc of center a and of radius r is $\varDelta (a,r)=\{z\in {\mathbb {D}}:\rho (a,z)<r\}$. It is well known that $\varDelta (a,r)$ is an Euclidean disk centered at $(1-r^2)a/(1-r^2|a|^2)$ and of radius $(1-|a|^2)r/(1-r^2|a|^2)$.

Theorem 2

Let $0<p\le q<\infty$, $\omega \in {\mathcal {D}}$ and $\mu$ a positive Borel measure on ${\mathbb {D}}$. Then there exists $r=r(\omega )\in (0,1)$ such that the following statements hold:

• (i) $\mu$ is a q-Carleson measure for $A^p_{\omega }$ if and only if

  $$\begin{aligned} \sup _{a\in {\mathbb {D}}}\frac{\mu \left( \varDelta (a,r)\right) }{\left( \omega \left( \varDelta (a,r)\right) \right) ^{\frac{q}{p}}}<\infty . \end{aligned}$$

  (7)

  Moreover, if $\mu$ is a q-Carleson measure for $A^p_{\omega }$, then the identity operator satisfies

  $$\begin{aligned} \Vert I_d\Vert ^q_{A^p_{\omega }\rightarrow L^q_{\mu }} \asymp \sup _{a\in {\mathbb {D}}} \frac{\mu \left( \varDelta (a,r)\right) }{\left( \omega \left( \varDelta (a,r)\right) \right) ^{\frac{q}{p}}}. \end{aligned}$$

• (ii) The identity operator $I_d:A^p_{\omega }\rightarrow L^q_{\mu }$ is compact if and only if

  $$\begin{aligned} \lim _{|a|\rightarrow 1^-}\frac{\mu \left( \varDelta (a,r)\right) }{\left( \omega \left( \varDelta (a,r) \right) \right) ^{\frac{q}{p}}}=0. \end{aligned}$$

  (8)

Another result needed is a lemma that allows us to estimate the distance between images of two points, say z and a, under f sufficiently accurately whenever z is close to a in the sense that $z\in \varDelta (a,r)$, and $f\in A^p_{\omega }$ with $\omega \in {\mathcal {D}}$. For the statement, denote $\widetilde{\omega }(z)=\widehat{\omega }(z)/(1-|z|)$ for all $z\in {\mathbb {D}}$.

Lemma 1

Let $\omega \in {\mathcal {D}}$, $0<p\le q<\infty$ and $0<r<R<1$. Then there exists a constant $C=C(\omega ,p,q,r,R)>0$ such that

$$\begin{aligned} |f(z)-f(a)|^q\le C\frac{\rho (z,a)^q}{\left( \widehat{\omega }(a)(1-|a|)\right) ^{\frac{q}{p}}} \int _{\varDelta (a,R)}|f(\zeta )|^p\widetilde{\omega }(\zeta )\,dA(\zeta ), \quad a\in {\mathbb {D}}, \quad z\in \varDelta (a,r), \end{aligned}$$

(9)

for all $f\in A^p_{\omega }$ with $\Vert f\Vert _{A^p_{\omega }}\le 1$.

This lemma plays an important role in the Proof of Theorem 1 when we show that (5) and (6) are sufficient conditions for the compactness. By [18, Proposition 5] we know that

$$\begin{aligned} \Vert f\Vert _{A^p_{\widetilde{\omega }}}\asymp \Vert f\Vert _{A^p_{\omega }},\quad f\in {\mathcal {H}}({\mathbb {D}}), \end{aligned}$$

(10)

provided $\omega \in {\mathcal {D}}$. This explains the appearance of the weight $\widetilde{\omega }$ on the right hand side of (9). It is worth observing that, despite of (10), the strictly positive weight $\widetilde{\omega }$ cannot be replaced by $\omega$ in the statement because $\omega \in {\mathcal {D}}$ may vanish in pseudohyperbolic discs of fixed radius that tend to the boundary.

The rest of the paper contains the proofs of the results stated above. We first prove Lemma 1 in the next section. The proof of the result on Carleson measures, Theorem 2, is given in Sect. 3, and finally, Theorem 1 is proved in Sect. 4.

To this end, couple of words about the notation used in the sequel. The letter $C=C(\cdot )$ will denote an absolute constant whose value depends on the parameters indicated in the parenthesis, and may change from one occurrence to another. We will use the notation $a\lesssim b$ if there exists a constant $C=C(\cdot )>0$ such that $a\le Cb$, and $a\gtrsim b$ is understood in an analogous manner. In particular, if $a\lesssim b$ and $a\gtrsim b$, then we write $a\asymp b$ and say that a and b are comparable.

2 Proof of Lemma 1

It is known that if $\omega \in {\mathcal {D}}$, then there exist constants $0<\alpha =\alpha (\omega )\le \beta (\omega )<\infty$ and $C=C(\omega )\ge 1$ such that

$$\begin{aligned} \frac{1}{C}\left( \frac{1-r}{1-t}\right) ^{\alpha } \le \frac{\widehat{\omega }(r)}{\widehat{\omega }(t)} \le C\left( \frac{1-r}{1-t}\right) ^{\beta },\quad 0\le r\le t<1. \end{aligned}$$

(11)

In fact, this pair of inequalities characterizes the class ${\mathcal {D}}$ because the right hand inequality is satisfied if and only if $\omega \in \widehat{{\mathcal {D}}}$ by [17, Lemma 2.1] while the left hand inequality describes the class $\check{{\mathcal {D}}}$ in an analogous way, see [14, (2.27)]. The chain of inequalities (11) will be frequently used in the sequel.

To prove the lemma, let $a\in {\mathbb {D}}$, $0<r<1$ and $z\in \varDelta (a,r)$. Then

$$\begin{aligned} |f(z)-f(a)|^p&=|f(\varphi _a(\varphi _a(z)))-f(\varphi _a(0))|^p =\left| \int _0^{\varphi _a(z)}(f\circ \varphi _a)'(\xi )\,d\xi \right| ^p\nonumber \\&\le \max _{\xi \in \overline{D(0,|\varphi _a(z)|)}}|(f\circ \varphi _a)'(\xi )|^p|\varphi _a(z)|^p \le \max _{\xi \in \overline{D(0,r)}}|(f\circ \varphi _a)'(\xi )|^p|\varphi _a(z)|^p. \end{aligned}$$

(12)

Let $R\in (r,1)$ and set $R'=\frac{r+R}{2}$. Further, let $0<s<1$. Then the Cauchy integral formula for the derivative and the subharmonicity of $|f|^p$ yield

$$\begin{aligned} |(f\circ \varphi _a)'(\xi )|^p&=\left| \frac{1}{2\pi }\int _{|w|=R'} \frac{(f\circ \varphi _a)(w)}{(w-\xi )^2}\,dw\right| ^p\le \left( \frac{2}{R-r} \right) ^{2p}\left( R'\right) ^p\max _{|w|=R'}|f(\varphi _a(w))|^p\nonumber \\&\lesssim \frac{4}{\pi s^2} \max _{|w|=R'}\frac{1}{(1-|\varphi _a(w)|)^2} \int _{\varDelta (\varphi _a(w),s)}|f(\zeta )|^p \,dA(\zeta )\nonumber \\&\lesssim \max _{|w|=R'}\frac{1}{(1-|a|)^2} \int _{\varDelta (\varphi _a(w),s)}| f(\zeta )|^p\,dA(\zeta ),\quad \xi \in \overline{D(0,r)}. \end{aligned}$$

(13)

Fix now $s=s(r,R)\in (0,1)$ sufficiently small such that $\varDelta (\varphi _a(w),s)\subset \varDelta (a,R)$ for all w such that $|w|=R'$. Further, an application of the right hand inequality in (11) shows that $\widehat{\omega }(\zeta )\asymp \widehat{\omega }(a)$ for all $\zeta \in \varDelta (a,R)$. Therefore, by combining (12) and (13) we deduce

$$\begin{aligned} |f(z)-f(a)|^p&\lesssim \frac{|\varphi _a(z)|^p}{(1-|a|)^2} \int _{\varDelta (a,R)}|f(\zeta )|^p\,dA(\zeta )\\&\lesssim \frac{|\varphi _a(z)|^p}{\widehat{\omega }(a)(1-|a|)} \int _{\varDelta (a,R)}|f(\zeta )|^p\,\widetilde{\omega }(\zeta ) \,dA(\zeta ),\quad a\in {\mathbb {D}},\quad z\in \varDelta (a,r). \end{aligned}$$

This proves the case $p=q$ because $|\varphi _a(z)|=\rho (z,a)$ for all $a,z\in {\mathbb {D}}$. This part of the proof is valid for all $f\in {\mathcal {H}}({\mathbb {D}})$ if $\omega \in \widehat{{\mathcal {D}}}$.

Let now $q>p$, and observe that trivially $|f(z)-f(a)|^q=\left( |f(z)-f(a)|^p\right) ^{\frac{q}{p}}$. An application of the case $q=p$ implies

$$\begin{aligned} |f(z)-f(a)|^q&\lesssim \frac{\rho (z,a)^q}{\left( \widehat{\omega }(a) (1-|a|)\right) ^{\frac{q}{p}}}\left( \int _{\varDelta (a,R)}| f(\zeta )|^p\widetilde{\omega }(\zeta )\,dA(\zeta )\right) ^{\frac{q}{p}}\\&\le \frac{\rho (z,a)^q\Vert f\Vert _{A^p_{\widetilde{\omega }}}^{q-p}}{\left( \widehat{\omega }(a)(1-|a|)\right) ^{\frac{q}{p}}} \int _{\varDelta (a,R)}|f(\zeta )|^p\widetilde{\omega }(\zeta )\,dA(\zeta ). \end{aligned}$$

But (10) guarantees $\Vert f\Vert _{A^p_{\widetilde{\omega }}}\asymp \Vert f\Vert _{A^p_{\omega }}\le 1$, and thus the assertion in the case $q>p$ follows from the above estimate.

3 Proof of Theorem 2

To prove (i), assume first (7) and let $0<r<1$. The fact that $|f|^p$ is subharmonic in ${\mathbb {D}}$ together with Minkowski’s inequality in continuous form (Fubini’s theorem in the case $q=p$) and (7) imply

$$\begin{aligned} \Vert f\Vert _{L^q_{\mu }}^q&\lesssim \int _{{\mathbb {D}}}\left( \int _{\varDelta (z,r)} \frac{|f(\zeta )|^p}{(1-|\zeta |)^2}\,dA(\zeta )\right) ^{\frac{q}{p}}\,d\mu (z)\\&\le \left( \int _{{\mathbb {D}}}|f(\zeta )|^p\frac{\mu (\varDelta (\zeta ,r))^{\frac{p}{q}}{(1-|\zeta |)^2}}{\,} dA(\zeta )\right) ^{\frac{q}{p}}\\&\lesssim \left( \int _{{\mathbb {D}}}|f(\zeta )|^p\frac{\omega (\varDelta (\zeta ,r))}{(1-|\zeta |)^2} \,dA(\zeta )\right) ^{\frac{q}{p}},\quad f\in {\mathcal {H}}({\mathbb {D}}). \end{aligned}$$

Since $\omega \in {\mathcal {D}}$ by the hypothesis, we may apply the right hand inequality in (11) to deduce

$$\begin{aligned} \omega (\varDelta (\zeta ,r))\lesssim \widehat{\omega }(\zeta )(1-|\zeta |),\quad \zeta \in {\mathbb {D}}. \end{aligned}$$

(14)

It follows that $\Vert f\Vert _{L^q_{\mu }}\lesssim \Vert f \Vert _{A^p_{\widetilde{\omega }}}$, and hence $\Vert f\Vert _{L^q_{\mu }} \lesssim \Vert f\Vert _{A^p_{\omega }}$ for all $f\in {\mathcal {H}}({\mathbb {D}})$ by (10). Thus $\mu$ is a q-Carleson measure $A^p_{\omega }$.

Conversely, assume that $\mu$ is a q-Carleson measure $A^p_{\omega }$. For each $a\in {\mathbb {D}}$, consider the function

$$\begin{aligned} f_{a}(z)=\left( \frac{1-|a|^2}{1-\overline{a}z}\right) ^{\gamma } \frac{1}{\left( \widehat{\omega }(a)(1-|a|)\right) ^{\frac{1}{p}}} =\frac{\left( 1-\overline{a}\varphi _a(z)\right) ^{\gamma }}{\left( \widehat{\omega } (a)(1-|a|)\right) ^{\frac{1}{p}}},\quad z\in {\mathbb {D}}, \end{aligned}$$

(15)

induced by $\omega$ and $0<\gamma ,p<\infty$. Then [17, Lemma 2.1] implies that for all $\gamma =\gamma (\omega ,p)>0$ sufficiently large we have $\Vert f_{a}\Vert _{A^p_{\omega }}\asymp 1$ for all $a\in {\mathbb {D}}$. Therefore the assumption yields

$$\begin{aligned} 1\asymp \Vert f_a\Vert ^q_{A^p_{\omega }}\gtrsim \Vert f_a\Vert _{L^q_{\mu }}^q \gtrsim \frac{\mu (\varDelta (a,r))}{(\widehat{\omega }(a)(1-|a|))^{\frac{q}{p}}},\quad a\in {\mathbb {D}}, \end{aligned}$$

that is, $\mu (\varDelta (a,r))\lesssim (\widehat{\omega }(a) (1-|a|))^{\frac{q}{p}}$ for all $a\in {\mathbb {D}}$. Since $\omega \in {\mathcal {D}}\subset \check{{\mathcal {D}}}$ by the hypothesis, there exists $K=K(\omega )>1$ and $C=C(\omega )>1$ such that $\widehat{\omega }(r)\ge C\widehat{\omega }\left( 1-\frac{1-r}{K} \right)$ for all $0\le r<1$ by the definition. Fix now $r=r(K)\in (0,1)$ sufficiently large such that

$$\begin{aligned} \varDelta (a,r)\supset \left\{ te^{i\theta }:|a|\le t \le 1-\frac{1-|a|}{K}, \,\,|\arg a-\theta |\le \frac{1-|a|}{2}\left( 1-\frac{1}{K}\right) \right\} . \end{aligned}$$

Then, as $\omega \in {\mathcal {D}}\subset \widehat{{\mathcal {D}}}$, the right hand inequality in (11) yields

$$\begin{aligned} \omega (\varDelta (a,r))&\ge (1-|a|)\left( 1-\frac{1}{K}\right) |a| \int _{|a|}^{1-\frac{1-|a|}{K}}\omega (s)\,ds\\&\ge (C-1)(1-|a|)\left( 1-\frac{1}{K}\right) |a|\widehat{\omega }\left( 1-\frac{1-|a|}{K}\right) \\&\gtrsim \widehat{\omega }(a)(1-|a|)|a|,\quad a\in {\mathbb {D}}, \end{aligned}$$

and therefore

$$\begin{aligned} \mu (\varDelta (a,r))\lesssim (\widehat{\omega }(a)(1-|a|))^{\frac{q}{p}} \lesssim \left( \frac{\omega (\varDelta (a,r))}{|a|}\right) ^{\frac{q}{p}},\quad a\in {\mathbb {D}}\setminus \{0\}. \end{aligned}$$

The claim (7) now follows from these estimates for all $r=r(\omega )\in (0,1)$ sufficiently large.

To prove (ii), assume first that $I_d:A^p_{\omega }\rightarrow L^q_{\mu }$ is compact. An application of [17, Lemma 2.1] and the right hand inequality in (11) ensure that we may choose $\gamma =\gamma (p,\omega )>0$ sufficiently large such that $\Vert f_{a}\Vert _{A^p_{\omega }}\asymp 1$ for all $a\in {\mathbb {D}}$, and $f_{a}\rightarrow 0$ uniformly on compact subsets of ${\mathbb {D}}$, as $|a|\rightarrow 1^-$. Therefore the closure of the set $\{f_{a}:a\in {\mathbb {D}}\}$ is compact in $L^q_{\mu }$. Since for each $\varepsilon >0$ the open balls $B(f_a,\varepsilon )=\{f\in L^q_{\mu }:\Vert f_a-f\Vert _{L^q_{\mu }}<\varepsilon \}$ cover $\overline{\{f_{a}:a\in {\mathbb {D}}\}}$, there exists a finite subcover $\left\{ B(f_{a_n},\varepsilon ):n=1,\ldots ,N=N(\varepsilon )\right\}$. Let now $a\in {\mathbb {D}}$ be arbitrary, and let $j=j(a)\in \{1,\ldots ,N\}$ such that $f_a\in B(f_{a_n},\varepsilon )$. Then, for each $R\in (0,1)$, we have

$$\begin{aligned} \int _{{\mathbb {D}}\setminus D(0,R)}|f_{a}(z)|^q\,d\mu (z)&\lesssim \int _{{\mathbb {D}}\setminus D(0,R)}|f_{a}(z)-f_{a_j}(z)|^q\,d\mu (z) +\int _{{\mathbb {D}}\setminus D(0,R)}|f_{a_j}(z)|^q\,d\mu (z)\\&\le \Vert f_a-f_{a_j}\Vert _{L^q_{\mu }}^q +\max _{n=1,\ldots ,N}\int _{{\mathbb {D}}\setminus D(0,R)}|f_{a_n}(z)|^q\,d\mu (z). \end{aligned}$$

By fixing $R\in (0,1)$ sufficiently large, and taking into account that $\varepsilon >0$ was arbitrary, we deduce

$$\begin{aligned} \lim _{R\rightarrow 1^-}\int _{{\mathbb {D}}\setminus D(0,R)}|f_{a}(z)|^q\,d\mu (z)=0 \end{aligned}$$

uniformly in a. This together with the uniform convergence yield

$$\begin{aligned} 0=\lim _{|a|\rightarrow 1^-}\Vert f_{a}\Vert _{L^q_{\mu }}^q \ge \lim _{|a|\rightarrow 1^-}\int _{\varDelta (a,r)}|f_{a}(z)|^q\,d\mu (z) \gtrsim \lim _{|a|\rightarrow 1^-}\frac{\mu (\varDelta (a,r))}{(\widehat{\omega }(a)(1-|a|))^{\frac{q}{p}}} \end{aligned}$$

for all $r\in (0,1)$. Now fix $r=r(\omega )$ as in the case (i) to have $\widehat{\omega }(a)(1-|a|)\lesssim \omega (\varDelta (a,r))$ for all $a\in {\mathbb {D}}$. Then we obtain (8).

Conversely, assume (8). Let $\{f_k\}_{k\in {\mathbb {N}}}$ be a sequence in $A^p_{\omega }$ such that $\sup _{k\in {\mathbb {N}}}\Vert f_k\Vert _{A^p_{\omega }} =M<\infty$. Then it is easy to see that $\{f_k\}_{k\in {\mathbb {N}}}$ is uniformly bounded on compact subsets of ${\mathbb {D}}$ – this follows, for example, from (18) below. Therefore $\{f_k\}_{k\in {\mathbb {N}}}$ constitutes a normal family by Montel’s theorem, and hence we may extract a subsequence $\{f_{k_j}\}_{j\in {\mathbb {N}}}$ that converges uniformly on compact subsets of ${\mathbb {D}}$ to a function f which belongs to ${\mathcal {H}}({\mathbb {D}})$ by Weierstrass’ theorem. Fatou’s lemma now shows that $f\in A^p_{\omega }$. For $r\in (0,1)$, fix an r-lattice $\{a_n\}_{n\in {\mathbb {N}}}$. Since $|a_n|\rightarrow 1$, as $n\rightarrow \infty$, we have

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{\mu (\varDelta (a_n,r))}{\omega (\varDelta (a_n,r))^{\frac{q}{p}}}=0 \end{aligned}$$

by the hypothesis. Therefore, for each $\varepsilon >0$, there exists $N=N(\varepsilon )\in {\mathbb {N}}$ such that

$$\begin{aligned} \frac{\mu (\varDelta (a_n,r))}{\omega (\varDelta (a_n,r))^{\frac{q}{p}}}<\varepsilon ,\quad n\ge N. \end{aligned}$$

Hence, as in the case (i), Minkowski’s inequality in continuous form (Fubini’s theorem in the case $q=p$), (14), (11) and (10) yield

$$\begin{aligned}&\sum _{n=N}^{\infty }\int _{\varDelta (a_n,r)}|f(z)-f_{k_j}(z)|^q\,d\mu (z)\\&\quad \lesssim \sum _{n=N}^{\infty }\int _{\varDelta (a_n,r)}\left( \int _{\varDelta (z,R)} \frac{|f(\zeta )-f_{k_j}(\zeta )|^p}{(1-|\zeta |)^2}\,dA(\zeta )\right) ^{\frac{q}{p}}\,d\mu (z)\\&\quad \le \sum _{n=N}^{\infty }\left( \int _{\{\zeta :\varDelta (a_n,r)\cap \varDelta (\zeta ,R) \ne \emptyset \}}\frac{|f(\zeta )-f_{k_j}(\zeta )|^p}{(1-|\zeta |)^2} \mu \left( \varDelta (a_n,r)\right) ^{\frac{p}{q}}dA(\zeta )\right) ^{\frac{q}{p}}\\&\quad \le \varepsilon \sum _{n=N}^{\infty }\left( \int _{\{\zeta :\varDelta (a_n,r)\cap \varDelta (\zeta ,R)\ne \emptyset \}}\frac{|f(\zeta )-f_{k_j}(\zeta )|^p}{(1-|\zeta |)^2} \omega \left( \varDelta (a_n,r)\right) dA(\zeta )\right) ^{\frac{q}{p}}\\&\quad \lesssim \varepsilon \left( \sum _{n=N}^{\infty }\int _{\{\zeta :\varDelta (a_n,r) \cap \varDelta (\zeta ,R)\ne \emptyset \}}|f(\zeta )-f_{k_j}(\zeta )|^p \widetilde{\omega }(\zeta )\,dA(\zeta )\right) ^{\frac{q}{p}}\\&\quad \lesssim \varepsilon \Vert f-f_{k_j}\Vert ^q_{A^p_{\widetilde{\omega }}} \asymp \varepsilon \Vert f-f_{k_j}\Vert ^q_{A^p_{\omega }} \lesssim M^q\varepsilon . \end{aligned}$$

Since

$$\begin{aligned} \lim _{j\rightarrow \infty }\sum _{n=1}^{N-1}\int _{\varDelta (a_n,r)}|f(z)-f_{k_j}(z)|^q\,d\mu (z)=0 \end{aligned}$$

by the uniform convergence in compact subsets, we deduce

$$\begin{aligned} \limsup _{j\rightarrow \infty }\int _{{\mathbb {D}}}|f(z)-f_{k_j}(z)|^q\,d\mu (z)&\le \limsup _{j\rightarrow \infty }\left( \sum _{n=1}^{N-1}\int _{\varDelta (a_n,r)} |f(z) -f_{k_j}(z)|^q\,d\mu (z)\right. \\&\quad \left. +\sum _{n=N}^{\infty }\int _{\varDelta (a_n,r)}|f(z) -f_{k_j}(z)|^q\,d\mu (z)\right) \lesssim \varepsilon . \end{aligned}$$

Since $\varepsilon >0$ was arbitrary, we have

$$\begin{aligned} \limsup _{j\rightarrow \infty }\int _{{\mathbb {D}}}|f(z)-f_{k_j}(z)|^q\,d\mu (z)=0, \end{aligned}$$

and hence $I_d:A^p_{\omega }\rightarrow L^q_{\mu }$ is compact. This completes the proof of the theorem.

4 Proof of Theorem 1

With the auxiliary results proved in the previous sections we are ready for the proof of the main result. We will follow the arguments used in [1] with appropriate modifications. The following two propositions will prove Theorem 1. The first one gives necessary conditions for $uC_{\varphi }-vC_{\psi }:A^p_{\omega } \rightarrow A^q_{\nu }$ to be compact.

Proposition 1

Let $\omega ,\nu \in {\mathcal {D}}$, $0<p, q<\infty$, $u,v\in {\mathcal {H}}({\mathbb {D}})$ and $\varphi$ and $\psi$ be analytic self-maps of ${\mathbb {D}}$. If $uC_{\varphi }-vC_{\psi }:A^p_{\omega }\rightarrow A^q_{\nu }$ is compact, then

$$\begin{aligned} \lim _{|z|\rightarrow 1^-}|\delta _1(z)|\left( |u(z)| \frac{\left( \widehat{\nu }(z)(1-|z|)\right) ^{\frac{1}{q}}}{\left( \widehat{\omega } (\varphi (z))(1-|\varphi (z)|)\right) ^{\frac{1}{p}}}+ |v(z)| \frac{\left( \widehat{\nu }(z)(1-|z|)\right) ^{\frac{1}{q}}}{\left( \widehat{\omega }(\psi (z))(1-|\psi (z)|)\right) ^{\frac{1}{p}}}\right) = 0 \end{aligned}$$

and there exits $\gamma =\gamma (\omega ,p)>0$ such that

$$\begin{aligned} \lim _{|z|\rightarrow 1^-}\left( \frac{|1-\overline{\varphi }(z)\delta _1(z)|^{\gamma }}{\left( \widehat{\omega }(\varphi (z))(1-|\varphi (z)|)\right) ^{\frac{1}{p}}} +\frac{|1-\overline{\psi }(z)\delta _2(z)|^{\gamma }}{\left( \widehat{\omega }(\psi (z)) (1-|\psi (z)|)\right) ^{\frac{1}{p}}}\right) |u(z)-v(z)|\left( \widehat{\nu }(z) (1-|z|)\right) ^{\frac{1}{q}}=0. \end{aligned}$$

Proof

Consider the test functions $f_a$ defined in (15), and set $F_{a}(z)=\varphi _a(z)f_{a}(z)$ for all $a,z\in {\mathbb {D}}$. Obviously, $\Vert F_{a}\Vert _{A^p_{\omega }}\le \Vert f_{a}\Vert _{A^p_{\omega }}$ for all $a\in {\mathbb {D}}$. Further, by the Proof of Theorem 2, both $f_{a}$ and $F_{a}$ tend to zero uniformly on compact subsets of ${\mathbb {D}}$ as $|a|\rightarrow 1^-$, and $\Vert f_{a}\Vert _{A^p_{\omega }}\asymp 1$ for all $a\in {\mathbb {D}}$ if $\gamma =\gamma (\omega ,p)>0$ is sufficiently large. Since $uC_{\varphi }-vC_{\psi }:A^p_{\omega } \rightarrow A^q_{\nu }$ is compact by the hypothesis, we therefore have

$$\begin{aligned} \lim _{|a|\rightarrow 1^-}\Vert uC_{\varphi }(f_{a})-vC_{\psi }(f_{a})\Vert _{A^q_{\nu }}=0 \end{aligned}$$

(16)

and

$$\begin{aligned} \lim _{|a|\rightarrow 1^-}\Vert uC_{\varphi }(F_{a})-vC_{\psi }(F_{a})\Vert _{A^q_{\nu }}=0. \end{aligned}$$

(17)

Also, if $\lim _{|a|\rightarrow 1^-}$ is replaced by $\sup _{a\in {\mathbb {D}}}$ in the above formulas, then the corresponding quantities are bounded.

We next observe that for each $\omega \in \widehat{{\mathcal {D}}}$ and $0<q<\infty$, there exists a positive bounded function $C_{\omega }$ on [0, 1) such that

$$\begin{aligned} |f(z)|\le \frac{C_{\omega }(|z|)}{\left( \widehat{\omega }(z)(1-|z|)\right) ^{\frac{1}{q}}} \Vert f\Vert _{A^q_{\omega }},\quad f\in A^q_{\omega },\quad z\in {\mathbb {D}}, \end{aligned}$$

(18)

and $C_{\omega }(z)\rightarrow 0$ as $|z|\rightarrow 1^-$. Namely, for each $z\in {\mathbb {D}}$ we have

$$\begin{aligned} \Vert f\Vert _{A^q_{\omega }}^q&\ge \int _{{\mathbb {D}}\setminus D(0,\frac{1+|z|}{2})}|f(\zeta )|^q \omega (\zeta )\,dA(\zeta ) \ge M_q^q\left( \frac{1+|z|}{2},f\right) \int _{\frac{1+|z|}{2}}^1r\omega (r)\,dr\\&\ge \frac{1}{2}M_q^q\left( \frac{1+|z|}{2},f\right) \widehat{\omega }\left( \frac{1+|z|}{2}\right) \gtrsim M_q^q\left( \frac{1+|z|}{2},f\right) \widehat{\omega }\left( z\right) , \end{aligned}$$

which combined with the well-known inequality $M_{\infty }(|z|,f) \lesssim M_q\left( \frac{1+|z|}{2},f\right) (1-|z|)^{-\frac{1}{q}}$, valid for all $f\in {\mathcal {H}}({\mathbb {D}})$, yields (18) because

$$\begin{aligned} \int _{{\mathbb {D}}\setminus D(0,\frac{1+|z|}{2})}|f(\zeta )|^q \omega (\zeta )\,dA(\zeta )\rightarrow 0,\quad |z|\rightarrow 1^-, \end{aligned}$$

for each $f\in A^q_{\omega }$.

By combining (16) and (17) with (18), we deduce

$$\begin{aligned} \lim _{\max (|a|,|z|)\rightarrow 1^-} |uC_{\varphi }(f_{a})(z)-vC_{\psi } (f_{a})(z)|\left( \widehat{\nu }(z)(1-|z|)\right) ^{\frac{1}{q}}=0 \end{aligned}$$

and

$$\begin{aligned} \lim _{\max (|a|,|z|)\rightarrow 1^-} |uC_{\varphi }(F_{a})(z)-vC_{\psi } (F_{a})(z)|\left( \widehat{\nu }(z)(1-|z|)\right) ^{\frac{1}{q}}=0. \end{aligned}$$

By choosing $a=\varphi (z)$, we obtain

$$\begin{aligned} \lim _{|z|\rightarrow 1^-} |u(z)f_{\varphi (z)}(\varphi (z))-v(z) f_{\varphi (z)}(\psi (z))|\left( \widehat{\nu }(z)(1-|z|)\right) ^{\frac{1}{q}}=0 \end{aligned}$$

(19)

and

$$\begin{aligned} \lim _{|z|\rightarrow 1^-} |\delta _1(z)||v(z)||f_{\varphi (z)} (\psi (z))|\left( \widehat{\nu }(z)(1-|z|)\right) ^{\frac{1}{q}}=0. \end{aligned}$$

(20)

Since $|\delta _1(z)|<1$ for all $z\in {\mathbb {D}}$, and

$$\begin{aligned} |u(z)||f_{\varphi (z)}(\varphi (z))| \le |u(z)f_{\varphi (z)}(\varphi (z)) -v(z)f_{\varphi (z)}(\psi (z))| +|v(z)||f_{\varphi (z)}(\psi (z))|, \end{aligned}$$

by combining (19) and (20) we deduce

$$\begin{aligned}&\lim _{|z|\rightarrow 1^-}\frac{|\delta _1(z)||u(z)|\left( \widehat{\nu }(z) (1-|z|)\right) ^{\frac{1}{q}}}{\left( \widehat{\omega }(\varphi (z)) (1-|\varphi (z)|)\right) ^{\frac{1}{p}}}\nonumber \\&\quad = \lim _{|z|\rightarrow 1^-}|\delta _1(z)||u(z)||f_{\varphi (z)} (\varphi (z))|\left( \widehat{\nu }(z)(1-|z|)\right) ^{\frac{1}{q}}=0. \end{aligned}$$

(21)

Further, we claim that for each $0<\gamma <\infty$ and each bounded set $\varOmega \subset {\mathbb {C}}$, there exists a constant $C=C(\gamma ,\varOmega )>0$ such that

$$\begin{aligned} |1-z^{\gamma }|\le C|1-z|,\quad z\in \varOmega , \end{aligned}$$

(22)

the proof of which is postponed for a moment. By using this and the fact that $1-\frac{1-|a|^2}{1-\overline{a}b}=\overline{a}\varphi _a(b)$ for all $a,b\in {\mathbb {D}}$, we deduce

$$\begin{aligned} |f_{\varphi (z)}(\varphi (z))-f_{\varphi (z)}(\psi (z))|&=|f_{\varphi (z)}(\varphi (z))|\left| 1-\left( \frac{1-|\varphi (z)|^2}{1-\overline{\varphi (z)}\psi (z)}\right) ^{\gamma }\right| \\&\lesssim |f_{\varphi (z)}(\varphi (z))|\left| 1-\frac{1-|\varphi (z)|^2}{1-\overline{\varphi (z)} \psi (z)}\right| \le |f_{\varphi (z)}(\varphi (z))||\delta _1(z)|, \end{aligned}$$

and hence

$$\begin{aligned} |u(z)-v(z)||f_{\varphi (z)}(\psi (z))|&\le |u(z)f_{\varphi (z)} (\varphi (z))-v(z)f_{\varphi (z)}(\psi (z))|\\&\quad +|u(z)||f_{\varphi (z)}(\varphi (z))-f_{\varphi (z)}(\psi (z))|\\&\lesssim |u(z)f_{\varphi (z)}(\varphi (z))-v(z)f_{\varphi (z)}(\psi (z))|\\&\quad +|\delta _1(z)||u(z)||f_{\varphi (z)}(\varphi (z))|,\quad z\in {\mathbb {D}}. \end{aligned}$$

Therefore, by (19) and (21), we finally obtain

$$\begin{aligned}&\lim _{|z|\rightarrow 1^-}\left( \frac{|1-\overline{\varphi }(z)\delta _1 (z)|^{\gamma }}{\left( \widehat{\omega }(\varphi (z))(1-|\varphi (z)|)\right) ^{\frac{1}{p}}}\right) |u(z)-v(z)|\left( \widehat{\nu }(z)(1-|z|)\right) ^{\frac{1}{q}}\\&=\lim _{|z|\rightarrow 1^-}|u(z)-v(z)||f_{\varphi (z)}(\psi (z))|\left( \widehat{\nu }(z) (1-|z|)\right) ^{\frac{1}{q}}=0. \end{aligned}$$

By following the reasoning above, but with the choice $a=\psi (z)$, we obtain

$$\begin{aligned} \lim _{|z|\rightarrow 1^-}\frac{|\delta _1(z)||v(z)|\left( \widehat{\nu } (z)(1-|z|)\right) ^{\frac{1}{q}}}{\left( \widehat{\omega }(\psi (z)) (1-|\psi (z)|)\right) ^{\frac{1}{p}}}=0 \end{aligned}$$

as an analogue of (21), and then eventually

$$\begin{aligned} \lim _{|z|\rightarrow 1^-}\left( \frac{|1-\overline{\psi }(z)\delta _2(z)|^{\gamma }}{\left( \widehat{\omega }(\psi (z))(1-|\psi (z)|)\right) ^{\frac{1}{p}}}\right) |u(z)-v(z) |\left( \widehat{\nu }(z)(1-|z|)\right) ^{\frac{1}{q}}=0. \end{aligned}$$

Therefore to finish the proof of the proposition, it remains to establish (22). If $z\in \varOmega \setminus \{z:|1-z| <1/2\}$, then

$$\begin{aligned} |1-z^{\gamma }|\le 1+\sup _{z\in \varOmega }|z|^{\gamma }\le 2 \left( 1+\sup _{z\in \varOmega }|z|^{\gamma }\right) |1-z|, \end{aligned}$$

while if $z\in \varOmega \cap \{z:|1-z|<1/2\}$, we have

$$\begin{aligned} |1-z^{\gamma }|=\left| \int _z^1\gamma \zeta ^{\gamma -1}\,d\zeta \right| \le \gamma \int _z^1|\zeta |^{\gamma -1}|d\zeta | \le \frac{\gamma \max \{1,3^{\gamma -1}\}}{2^{\gamma -1}}|1-z|. \end{aligned}$$

This proves (22), and completes the proof of the proposition. $\square$

Sufficient conditions for the compactness of $uC_{\varphi } -vC_{\psi }:A^p_{\omega }\rightarrow A^q_{\nu }$ are given in the next result.

Proposition 2

Let $\omega ,\nu \in {\mathcal {D}}$ and $0<p\le q<\infty$ such that $\widehat{\nu }(z)(1-|z|)\lesssim \left( \widehat{\omega }(z) (1-|z|)\right) ^{\frac{q}{p}}$ for all $z\in {\mathbb {D}}$. Further, let $u,v\in {\mathcal {H}}({\mathbb {D}})$ and $\varphi$ and $\psi$ analytic self-maps of ${\mathbb {D}}$ such that (4) is satisfied. If there exists $\gamma >0$ such that (5) and (6) are satisfied, then $uC_{\varphi }-vC_{\psi }:A^p_{\omega }\rightarrow A^q_{\nu }$ is compact.

Proof

It suffices to show that for any norm bounded sequence $\{f_n\}$ in $A^p_{\omega }$ which tends to zero uniformly on compact subsets of ${\mathbb {D}}$ as $n\rightarrow \infty$, we have $\Vert (uC_{\varphi }-vC_{\psi })(f_n) \Vert _{A^q_{\nu }}\rightarrow 0$ as $n\rightarrow \infty$. For simplicity, assume $\Vert f_n\Vert _{A^p_{\omega }}\le 1$ for all n. Fix $0<r<R<1$, and denote $E=\{z\in {\mathbb {D}}:|\delta _1(z)|<r\}$ and $E'={\mathbb {D}}\setminus E$. Write

$$\begin{aligned} (uC_{\varphi }-vC_{\psi })(f_n)=(uC_{\varphi }-vC_{\psi })(f_n)\chi _{E'}+(u-v)C_{\psi } (f_n)\chi _E+u(C_{\varphi }-C_{\psi })(f_n)\chi _E, \end{aligned}$$

and observe that it is enough to prove that each of the three quantities

$$\begin{aligned} \Vert (uC_{\varphi }-vC_{\psi })(f_n)\chi _{E'}\Vert _{ A^q_{\nu }}, \quad \Vert (u-v)C_{\psi }(f_n)\chi _E\Vert _{ A^q_{\nu }}\quad {\text {and}} \quad \Vert u(C_{\varphi }-C_{\psi })(f_n)\chi _E\Vert _{ A^q_{\nu }} \end{aligned}$$

(23)

tends to zero as $n\rightarrow \infty$.

We begin with considering the first two quantities in (23). By the definition of the set E we have the estimates

$$\begin{aligned} |(uC_{\varphi }-vC_{\psi })(f_n)\chi _{E'}|\le \frac{1}{r}(|\delta _1 uC_{\varphi } (f_n)|+|\delta _1vC_{\psi }(f_n)|) \end{aligned}$$

and

$$\begin{aligned} |(u-v)C_{\psi }(f_n)\chi _E|\le \left( \frac{1}{1-r}\right) ^{\gamma }|1 -\overline{\psi }\delta _2|^{\gamma }|u-v||C_{\psi }(f_n)| \end{aligned}$$

on ${\mathbb {D}}$. Therefore it suffices to prove that $\delta _1uC_{\varphi }$, $\delta _1vC_{\psi }$ and $(1-\overline{\psi }\delta _2)^{\gamma }(u-v) C_{\psi }$ are compact operators from $A^p_{\omega }$ to $L^q_{\nu }$. We show in detail that $\delta _1uC_{\varphi }$ is compact - the same argument shows the compactness of the other two operators.

Let $\mu$ be a finite nonnegative Borel measure on ${\mathbb {D}}$ and h a measureable function on ${\mathbb {D}}$. For an analytic self-map $\varphi$ of ${\mathbb {D}}$, the weighted pushforward measure is defined by

$$\begin{aligned} \varphi _*(h,\mu )(M)=\int _{\varphi ^{-1}(M)}h d\mu \end{aligned}$$

(24)

for each measurable set $M\subset {\mathbb {D}}$. If $\mu$ is the Lebesgue measure, we omit the measure in the notation and write $\varphi _*(h)(M)$ for the left hand side of (24). By the measure theoretic change of variable [9, Section 39], we have $\Vert \delta _1uC_{\varphi }(f)\Vert _{L^q_{\nu }}=\Vert f\Vert _{L^q_{\varphi _* (|\delta _1u|^q\nu )}}$ for each $f\in A^p_{\omega }$. Therefore Theorem 2 shows that $\delta _1 uC_{\varphi }:A^p_{\omega }\rightarrow L^q_{\nu }$ is compact if and only if

$$\begin{aligned} \frac{\varphi _*(|\delta _1u|^q\nu )(\varDelta (a,r))}{\omega (\varDelta (a,r))^{\frac{q}{p}}} =\frac{\int _{\varphi ^{-1}(\varDelta (a,r))}|\delta _1(z)u(z)|^q\nu (z)\,dA(z)}{\omega (\varDelta (a,r))^{\frac{q}{p}}}\rightarrow 0,\quad |a|\rightarrow 1^-. \end{aligned}$$

This is what we prove next. Define

$$\begin{aligned} W_{a,r}=\sup _{z\in \varphi ^{-1}(\varDelta (a,r))}\left| \left( \delta _1(z)u(z)\right) ^q \frac{\widehat{\nu }(z)(1-|z|)}{\left( \widehat{\omega }(\varphi (z)) (1-|\varphi (z)|)\right) ^{\frac{q}{p}}}\right| . \end{aligned}$$

Then $W_{a,r}\rightarrow 0$, as $|a|\rightarrow 1^-$, by the hypothesis (5). Moreover, for $a\in {\mathbb {D}}$ and $z\in \varphi ^{-1}(\varDelta (a,r))$, (11) yields

$$\begin{aligned} \left| \delta _1(z)u(z)\right| ^q \lesssim W_{a,r}\frac{\left( \widehat{\omega }(a) (1-|a|)\right) ^{\frac{q}{p}}}{\widehat{\nu }(z)(1-|z|)}, \end{aligned}$$

and therefore, for each $\varepsilon \in (0,1)$ we have

$$\begin{aligned} \varphi _*(|\delta _1u|^q\nu )(\varDelta (a,r))&=\int _{\varphi ^{-1}(\varDelta (a,r))} \left| \delta _1(z)u(z)\right| ^q\nu (z)\,dA(z)\nonumber \\&\lesssim \sup _{z\in {\mathbb {D}}}\left| \left( \delta _1(z)u(z)\right) ^q \frac{\widehat{\nu }(z)(1-|z|)}{\left( \widehat{\omega }(z)(1-|z|)\right) ^{\frac{q}{p}}}\right| ^{1-\varepsilon } W_{a,r}^{\varepsilon } \left( \widehat{\omega }(a)(1-|a|)\right) ^{{\varepsilon \frac{q}{p}}} \nonumber \\&\quad \cdot \int _{\varphi ^{-1}(\varDelta (a,r))}\frac{\left( \widehat{\omega }(z) (1-|z|)\right) ^{(1-\varepsilon )\frac{q}{p}}\nu (z)}{\widehat{\nu }(z)(1-|z|)}dA(z),\quad a\in {\mathbb {D}}. \end{aligned}$$

(25)

Before proceeding further, we indicate how to get to this point with the operator $(1-\overline{\psi }\delta _2)^{\gamma }(u-v)C_{\psi }$. After the measure theoretic change of variable and an application of Theorem 2, consider

$$\begin{aligned} V_{a,r}= \sup _{z\in \psi ^{-1}(\varDelta (a,r))} \left| \left( 1-\overline{\psi }(z) \delta _2(z)\right) ^{\gamma q} \left( u(z)-v(z)\right) ^q\frac{\widehat{\nu }(z)(1-|z|)}{\left( \widehat{\omega }(\psi (z))(1-|\psi (z)|)\right) ^{\frac{q}{p}}}\right| \end{aligned}$$

instead of $W_{a,r}$. Then $V_{a,r}\rightarrow 0$, as $|a|\rightarrow 1^-$, by the hypothesis (6), and moreover, for $a\in {\mathbb {D}}$ and $z\in \psi ^{-1}(\varDelta (a,r))$, (11) yields

$$\begin{aligned} \left| 1-\overline{\psi }(z)\delta _2(z)\right| ^{\gamma q} \left| u(z)-v(z)\right| ^q \lesssim V_{a,r}\frac{\left( \widehat{\omega }(a)(1-|a|)\right) ^{\frac{q}{p}}}{\widehat{\nu }(z)(1-|z|)}. \end{aligned}$$

Therefore, for each $\varepsilon \in (0,1)$, we have

$$\begin{aligned}&\psi _*(\left| 1-\overline{\psi }\delta _2\right| ^{\gamma q}|u-v|^q\nu )(\varDelta (a,r))\\&\lesssim \sup _{z\in {\mathbb {D}}} \left| \left( 1-\overline{\psi }(z)\delta _2(z)\right) ^{\gamma q} \left( u(z)-v(z)\right) ^q \frac{\widehat{\nu }(z)(1-|z|)}{\left( \widehat{\omega }(z)(1-|z|)\right) ^{\frac{q}{p}}}\right| ^{1-\varepsilon } V_{a,r}^{\varepsilon } \left( \widehat{\omega }(a)(1-|a|)\right) ^{{\varepsilon \frac{q}{p}}} \\&\quad \cdot \int _{\psi ^{-1}(\varDelta (a,r))}\frac{\left( \widehat{\omega }(z) (1-|z|)\right) ^{(1-\varepsilon )\frac{q}{p}}\nu (z)}{\widehat{\nu }(z)(1-|z|)}dA(z),\quad a\in {\mathbb {D}}. \end{aligned}$$

To estimate this last integral, which is the same as the one appearing in (25), we first show that there exists $\varepsilon =\varepsilon (\omega ,\nu ,q,p)\in (0,1)$ sufficiently small such that the function

$$\begin{aligned} \mu (z)=\mu _{\omega ,\nu ,\varepsilon ,q,p}(z) =\frac{\left( \widehat{\omega }(z)(1-|z|)\right) ^{(1-\varepsilon ) \frac{q}{p}}\nu (z)}{\widehat{\nu }(z)(1-|z|)},\quad z\in {\mathbb {D}}, \end{aligned}$$

is a weight and belongs to ${\mathcal {D}}$. To see that $\mu \in \widehat{{\mathcal {D}}}$, for $n\in {\mathbb {N}}\cup \{0\}$, define $\rho _n$ by $\widehat{\nu }(\rho _n) =\frac{\widehat{\nu }(0)}{K^n}$, where $K>1$. Let $0\le r<1$ and fix $M\in {\mathbb {N}}\cup \{0\}$ such that $\rho _M\le r<\rho _{M+1}$. Set $\alpha =(1-\varepsilon )\frac{q}{p}$ for short. Since $\omega ,\nu \in {\mathcal {D}}$ by the hypothesis, (11) yields

$$\begin{aligned} \widehat{\omega }(\rho _{j+N+1})&\le \widehat{\omega }(\rho _{j}) \lesssim \widehat{\omega }(\rho _{j+N+1})\left( \frac{1-\rho _j}{1-\rho _{j+N+1}} \right) ^{\beta (\omega )} \lesssim \widehat{\omega }(\rho _{j+N+1}) \left( \frac{\widehat{\nu }(\rho _{j})}{\widehat{\nu } (\rho _{j+N+1})}\right) ^{\frac{\beta (\omega )}{\alpha (\nu )}}\\&=\widehat{\omega }(\rho _{j+N+1})K^{(N+1)\frac{\beta (\omega )}{\alpha (\nu )}} \asymp \widehat{\omega }(\rho _{j+N+1}) \end{aligned}$$

and $1-\rho _j\asymp 1-\rho _{j+1}$ for all j and for each fixed $N\in {\mathbb {N}}$. Therefore

$$\begin{aligned} \int _r^1\frac{(\widehat{\omega }(t))^{\alpha }}{(1-t)^{1-\alpha }} \frac{\nu (t)}{\widehat{\nu }(t)}dt&\le \sum _{j=M}^{\infty } \int _{\rho _j}^{\rho _{j+1}}\frac{(\widehat{\omega }(t))^{\alpha }}{(1-t)^{1-\alpha }}\frac{\nu (t)}{\widehat{\nu }(t)}\,dt \le \sum _{j=M}^{\infty }\frac{(\widehat{\omega }(\rho _j))^{\alpha }}{(1-\rho _{j+1})^{1-\alpha }} \int _{\rho _{j}}^{\rho _{j+1}}\frac{\nu (t)}{\widehat{\nu }(t)}dt\\&=\sum _{j=M}^{\infty }\frac{(\widehat{\omega }(\rho _j))^{\alpha }}{(1-\rho _{j+1})^{1-\alpha }} \int _{\rho _{j+N}}^{\rho _{j+N+1}}\frac{\nu (t)}{\widehat{\nu }(t)}dt \asymp \int _{\rho _{M+N}}^1\frac{(\widehat{\omega }(t))^{\alpha }}{(1-t)^{1-\alpha }} \frac{\nu (t)}{\widehat{\nu }(t)}\,dt. \end{aligned}$$

Another application of (11) shows that there exists $N=N(\nu )\in {\mathbb {N}}$ such that $\rho _{M+N}\ge \frac{1+\rho _{M+1}}{2}$. Namely, the right hand inequality implies

$$\begin{aligned} \frac{1-\rho _{M+1}}{1-\rho _{M+N}}\ge \left( \frac{\widehat{\nu }(\rho _{M+1})}{C\widehat{\nu } (\rho _{M+N})}\right) ^{\frac{1}{\beta }} =\left( \frac{K^{N-1}}{C}\right) ^{\frac{1}{\beta }}\ge 2 \end{aligned}$$

for sufficiently large N giving what we want. Therefore

$$\begin{aligned} \int _r^1\frac{(\widehat{\omega }(t))^{\alpha }}{(1-t)^{1-\alpha }}\frac{\nu (t)}{\widehat{\nu }(t)}dt \lesssim \int _{{\frac{1+\rho _{M+1}}{2}}}^1\frac{(\widehat{\omega }(t))^{\alpha }}{(1-t)^{1-\alpha }} \frac{\nu (t)}{\widehat{\nu }(t)}dt \le \int _{{\frac{1+r}{2}}}^1 \frac{(\widehat{\omega }(t))^{\alpha }}{(1-t)^{1-\alpha }}\frac{\nu (t)}{\widehat{\nu }(t)}dt, \end{aligned}$$

which shows that $\mu \in \widehat{{\mathcal {D}}}$, provided $\mu$ is a weight.

We next show that $\mu$ is a weight in $\check{{\mathcal {D}}}$. By using (11), with $\beta$ in place of $\alpha$, we obtain

$$\begin{aligned} \int _r^1\frac{(\widehat{\omega }(t))^{\alpha }}{(1-t)^{1-\alpha }}\frac{\nu (t)}{\widehat{\nu }(t)}dt&\lesssim \int _r^1 \frac{\left( \widehat{\omega }(r)\left( \frac{1-t}{1-r}\right) ^{\beta } (1-t)\right) ^{\alpha }}{\widehat{\nu }(t)(1-t)}\nu (t)\,dt\\&\lesssim \frac{\widehat{\omega }(r)^{\alpha }}{(1-r)^{\beta \alpha }} \int _r^1 \frac{(1-t)^{{(1+\beta )\alpha }}}{\widehat{\nu }(t)(1-t)}\nu (t)\,dt. \end{aligned}$$

Now fix $\varepsilon \in (0,1)$ sufficient small such that $\sigma =(1+\beta )\alpha =(1+\beta )(1-\varepsilon )\frac{q}{p}>1$. By [19, Lemma 3], the last expression above is dominated by a constant times

$$\begin{aligned} \frac{\widehat{\omega }(r)^{\alpha }}{(1-r)^{\beta \alpha }}(1-r)^{\sigma -1} =\frac{\left( \widehat{\omega }(r)(1-r)\right) ^{\alpha }}{1-r}. \end{aligned}$$

Since $\nu \in \check{{\mathcal {D}}}$ by the hypothesis, there exists $K=K(\nu )>1$ such that

$$\begin{aligned} \frac{\left( \widehat{\omega }(r)(1-r)\right) ^{\alpha }}{1-r}&\asymp \frac{\left( \widehat{\omega }(r)(1-r)\right) ^{\alpha }}{\widehat{\nu }(r) (1-r)}\int _r^{1-\frac{1-r}{K}}\nu (t)dt\\&\asymp \int _r^{1-\frac{1-r}{K}}\frac{\left( \widehat{\omega }(t)(1-t)\right) ^{\alpha }}{\widehat{\nu }(t)(1-t)} \nu (t)dt, \end{aligned}$$

where the last step is a consequence of (11), applied to both weights $\omega ,\nu \in {\mathcal {D}}$. This reasoning shows that $\mu$ is a weight in $\check{{\mathcal {D}}}$, and thus $\mu \in {\mathcal {D}}$.

We return to estimate the last integral in (25). By [16, Proposition 18], the operator $C_{\varphi }:A^p_{\mu }\rightarrow A^p_{\mu }$ is bounded for each $0<p<\infty$. By the measure theoretic change of variable, this is equivalent to saying that $I_d:A^p_{\mu }\rightarrow L^p_{\varphi _{*}(\mu )}$ is bounded. Since we just proved that $\mu \in {\mathcal {D}}$, this is in turn equivalent to $\varphi _{*}(\mu )(\varDelta (a,r))\lesssim \mu (\varDelta (a,r))$ by Theorem 2. By the definition of these two measures and the poof of the fact $\mu \in \check{{\mathcal {D}}}$ above, we have

$$\begin{aligned} \int _{\varphi ^{-1}(\varDelta (a,r))}\mu (z)\,dA(z) \lesssim \int _{\varDelta (a,r)} \mu (z)\,dA(z) \lesssim \left( \widehat{\omega }(a)(1-|a|)\right) ^{{(1-\varepsilon ) \frac{q}{p}}},\quad a\in {\mathbb {D}}. \end{aligned}$$

This combined with (25) gives

$$\begin{aligned} \varphi _*(|\delta _1u|^q\nu )(\varDelta (a,r)) \lesssim \sup _{z\in {\mathbb {D}}} \left| \left( \delta _1(z)u(z)\right) ^q \frac{\widehat{\nu }(z)(1-|z|)}{\left( \widehat{\omega }(z)(1-|z|)\right) ^{\frac{q}{p}}}\right| ^{1-\varepsilon } W_{a,r}^{\varepsilon } \left( \widehat{\omega }(a)(1-|a|)\right) ^{{\frac{q}{p}}}. \end{aligned}$$

Since the supremum above is bounded by the hypothesis (4), and $\widehat{\omega }(a)(1-|a|)\lesssim \omega (\varDelta (a,r))$ for $r=r(\omega )\in (0,1)$ sufficiently large by the Proof of Theorem 2, we deduce via Theorem 2 that $\delta _1uC_{\varphi }:A^p_{\omega }\rightarrow L^q_{\nu }$ is compact. As mentioned already, $\delta _1vC_{\psi }$ and $(1-\overline{\psi }\delta _2)^{\gamma }(u-v)C_{\psi }$ can be treated in the same way.

It remains to deal with the third term in (23). By Lemma 1, Fubini’s theorem and (11), we have

$$\begin{aligned} \Vert u(C_{\varphi }-C_{\psi })(f_n)\chi _E\Vert ^q_{A^q_{\nu }}&=\int _E|u(z)|^q|f_n(\varphi (z))-f_n(\psi (z))|^q\nu (z)\,dA(z)\\&\lesssim \int _E\frac{|u(z)\delta _1(z)|^q}{\left( \widehat{\omega } (\varphi (z))(1-|\varphi (z)|)\right) ^{\frac{q}{p}}} \int _{(\varphi (z),R)}|f_n(\zeta )|^p\widetilde{\omega }(\zeta )\,dA(\zeta )\nu (z)\,dA(z)\\&\le \int _{{\mathbb {D}}}|f_n(\zeta )|^p\widetilde{\omega }(\zeta )\\&\quad \cdot \left( \int _{\varphi ^{-1}(\varDelta (\zeta ,R))\cap E} \frac{|u(z)\delta _1(z)|^q}{\left( \widehat{\omega }(\varphi (z)) (1-|\varphi (z)|)\right) ^{\frac{q}{p}}}\nu (z)\,dA(z)\right) dA(\zeta )\\&\asymp \int _{{\mathbb {D}}}|f_n(\zeta )|^p\widetilde{\omega }(\zeta )\left( \int _{\varphi ^{-1} (\varDelta (\zeta ,R))}\frac{|u(z)\delta _1(z)|^q}{\left( \widehat{\omega }(\zeta )(1-|\zeta |)\right) ^{\frac{q}{p}}} \nu (z) \,dA(z)\right) dA(\zeta )\\&=\int _{{\mathbb {D}}}|f_n(\zeta )|^p\frac{\varphi _*(|\delta _1 u|^q\nu )(\varDelta (\zeta ,R))}{\left( \widehat{\omega }(\zeta )(1-|\zeta |) \right) ^{\frac{q}{p}}}\widetilde{\omega }(\zeta )\,dA(\zeta ). \end{aligned}$$

Since the identity operator from $A^p_{\omega }$ to $L^q_{\varphi _*(|\delta _1u|^q\nu )}$ is compact, it is also bounded. This and Theorem 2 yield

$$\begin{aligned} \Vert u(C_{\varphi }-C_{\psi })(f_n)\chi _E\Vert ^q_{A^q_{\nu }}&\lesssim \sup _{\zeta \in D(0,r)} \frac{\varphi _*(|\delta _1u|^q\nu )(\varDelta (\zeta ,R))}{\left( \widehat{\omega }(\zeta )(1-|\zeta |) \right) ^{\frac{q}{p}}}\int _{D(0,r)}|f_n(\zeta )|^p\widetilde{\omega }(\zeta )\,dA(\zeta )\\&\quad +\sup _{\zeta \in {\mathbb {D}}\setminus D(0,r)}\frac{\varphi _*(|\delta _1u|^q\nu ) (\varDelta (\zeta ,R))}{\left( \widehat{\omega } (\zeta )(1-|\zeta |)\right) ^{\frac{q}{p}}} \int _{{\mathbb {D}}\setminus D(0,r)}|f_n(\zeta )|^p\widetilde{\omega }(\zeta )\,dA(\zeta )\\&\lesssim \sup _{\zeta \in D(0,r)}|f_n(\zeta )|^p + \sup _{\zeta \in {\mathbb {D}}\setminus D(0,r)} \frac{\varphi _*(|\delta _1u|^q\nu ) (\varDelta (\zeta ,R))}{\left( \widehat{\omega } (\zeta )(1-|\zeta |)\right) ^{\frac{q}{p}}},\quad 0<r<1. \end{aligned}$$

By choosing $0<r<1$ sufficiently large, the last term can be made smaller than a pregiven $\varepsilon >0$. For such fixed r, the first term tends to zero as $n\rightarrow \infty$ by the uniform convergence. Therefore

$$\begin{aligned} \Vert u(C_{\varphi }-C_{\psi })(f_n)\chi _E\Vert ^q_{A^q_{\nu }}\rightarrow 0,\quad n\rightarrow \infty , \end{aligned}$$

and hence also the last term in (23) tends to zero. This finishes the proof of the proposition. $\square$