# UG

and  Jean$-$Marcel Tanoh Dje, Felix Ofori and Benoît F. Sehba  Unité de
Recherche et d’Expertise Numérique, Université virtuelle de Côte d’Ivoire,
Abidjan<EMAIL_ADDRESS>Department of Mathematics, University of
Ghana, P.O. Box L.G 62 Legon, Accra, Ghana<EMAIL_ADDRESS>Department
of Mathematics, University of Ghana, P.O. Box L.G 62 Legon, Accra, Ghana.
<EMAIL_ADDRESS>
# Boundedness of the Bergman projection on some weighted mixed norm Lebesgue
spaces of the upper-half space

and  Jean$-$Marcel Tanoh Dje, Felix Ofori and Benoît F. Sehba  Unité de
Recherche et d’Expertise Numérique, Université virtuelle de Côte d’Ivoire,
Abidjan<EMAIL_ADDRESS>Department of Mathematics, University of
Ghana, P.O. Box L.G 62 Legon, Accra, Ghana<EMAIL_ADDRESS>Department
of Mathematics, University of Ghana, P.O. Box L.G 62 Legon, Accra, Ghana.
<EMAIL_ADDRESS>
###### Abstract.

In this paper, we prove the boundedness of the Bergman projection on weighted
mixed norm spaces of the upper-half space for some weights that are
constructed using the logarithm function and growth functions. Our necessary
and sufficient condition is the same as in the unweighted case, that is it
involves only the parameters and not the weight. We then provide some
applications in terms of dual and derivative characterization, and an atomic
decomposition of the corresponding Bergman spaces.

###### Key words and phrases:

Bergman space, Hardy space, Bergman projection, Atomic decomposition, weight

###### 1991 Mathematics Subject Classification:

30H20, 32A25,42B35

##  1\. Introduction

We put ourselves in the upper-half space
${\mathbb{C}}_{+}:=\\{z\in{\mathbb{C}}:\Im mz>0\\}$. We denote by
$\mathcal{H}({\mathbb{C}}_{+})$ the set of holomorphic functions in
${\mathbb{C}}_{+}$. Let $\alpha>-1$, and let $1\leq p,q<\infty$. For $\Omega$
a weight (positive function) on $(0,\infty)$, we denote by
$L_{\Omega,\alpha}^{p,q}({\mathbb{C}}_{+})$ the set of all measurable
functions $f$ on $\mathbb{C}_{+}$ such that

$\|f\|_{\Omega,\alpha,p,q}^{q}:=\int_{0}^{\infty}\left(\int_{\mathbb{R}}|f(x+iy)|^{p}dx\right)^{\frac{q}{p}}\Omega(y)y^{\alpha}dy<\infty.$

In particular, when $\Omega(y)=1$ and $\alpha>-1$, we use the notations
$L_{\alpha}^{p,q}(\mathbb{C}_{+})=L^{p,q}(\mathbb{C}_{+},(\Im
mz)^{\alpha}dV(z))$ for the corresponding space and $\|\cdot\|_{\alpha,p,q}$
for the corresponding norm.

For $1\leq p,q<\infty$ and $\alpha>-1$, the mixed norm Bergman space
$A_{\Omega,\alpha}^{p,q}(\mathbb{C}_{+})$ is the intersection
$L_{\Omega,\alpha}^{p,q}(\mathbb{C}_{+})\cap\mathcal{H}({\mathbb{C}}_{+})$. We
note that depending on the weight, this space can be trivial.

When $p=q$, the above space is just denoted
$L_{\Omega,\alpha}^{p}({\mathbb{C}}_{+})$, and its norm
$\|\cdot\|_{\Omega,\alpha,p}$.

For simplicity, we will be using the notation

$dV_{\alpha}(z)=\left(\Im
m(z)\right)^{\alpha}dV(z)=y^{\alpha}dxdy\quad\textrm{for}\quad z=x+iy.$

For $\alpha>-1$, the Bergman projection $P_{\alpha}$ is the orthogonal
projection from $L_{\alpha}^{2}(\mathbb{C}_{+})$ into its closed subspace
$A_{\alpha}^{2}(\mathbb{C}_{+})$. This operator is given by

$P_{\alpha}f(z)=\int_{\mathbb{C}_{+}}K_{\alpha}(z,w)f(w)dV_{\alpha}(w)$

where

$K_{\alpha}(z,w)=\frac{c_{\alpha}}{\left(z-\bar{w}\right)^{2+\alpha}}$

is the Bergman kernel, and $c_{\alpha}$ is a constant that depends only on
$\alpha$. We denote by $P_{\alpha}^{+}$, the operator with kernel
$\left|K_{\alpha}(\cdot,\cdot)\right|$.

It is a well known and easy to verify fact that the Bergman projection
$P_{\alpha}$ (and equivalently $P_{\alpha}^{+}$) extends to a bounded
projection on $L_{\alpha}^{p}(\mathbb{C}_{+})$ for any $1<p<\infty$. The
problem of characterizing the weights $\omega$ such that $P_{\alpha}$ is
bounded on $L^{p}(\mathbb{C}_{+},\omega(z)dV_{\alpha}(z))$ was considered and
solved by D. Békollé and A. Bonami in 1978 and 1982 (see [2, 3]). They proved
that a necessary and sufficient condition for this to hold, is that the weight
$\omega$ should satisfy the following condition also known as the
$B_{p}$-condition:

(1.1)
$[\omega]_{p}:=\sup_{I}\left(\frac{1}{|Q_{I}|_{\alpha}}\int_{Q_{I}}\omega(z)dV_{\alpha}(z)\right)\left(\frac{1}{|Q_{I}|_{\alpha}}\int_{Q_{I}}\left(\omega(z)\right)^{-\frac{p^{\prime}}{p}}dV_{\alpha}(z)\right)^{\frac{p}{p^{\prime}}}<\infty.$

Here the supremum is taken over all subintervals $I$ of $\mathbb{R}$, $Q_{I}$
is the square above $I$, and $pp^{\prime}=p^{\prime}+p$.

Obviously, this condition is difficult to verify for general weights.
Moreover, this result doesn’t say anything about the case of mixed norm spaces
defined above.

Our aim in this paper is to give an answer to the problem of boundedness of
the Bergman projection on weighted mixed norm spaces for an appropriate family
of weights on $(0,\infty)$. The particularity here is that for the weights
considered here, the condition obtained is the same as in the unweighted case
(see [1]).

Let us defined our weights. We recall that a growth function is any function
$\Phi:\mathbb{R}_{+}\longrightarrow\mathbb{R}_{+}$, not identically null, that
is continuous, nondecreasing and onto. We observe that if $\Phi$ is a growth
function, then $\Phi(0)=0$ and $\lim_{t\to\infty}\Phi(t)=\infty$.

Let $p>0$ be a real and $\Phi$ a growth function. We say that $\Phi$ is of
upper-type (resp. lower-type) $p>0$ if there exists a constant $C_{p}>0$ such
that for all $t\geq 1$ (resp. $0<t\leq 1$),

(1.2) $\Phi(st)\leq C_{p}t^{p}\Phi(s),~{}~{}\forall~{}s>0.$

We denote by $\mathscr{U}^{p}$ (resp. $\mathscr{L}_{p}$) the set of all growth
functions of upper-type $p\geq 1$ (resp. lower-type $0<p\leq 1$) such that the
function $t\mapsto\frac{\Phi(t)}{t}$ is non decreasing (resp. non-increasing)
on $\mathbb{R}_{+}^{*}=\mathbb{R}_{+}\backslash\\{0\\}$. We put
$\mathscr{U}:=\bigcup_{p\geq 1}\mathscr{U}^{p}$ (resp.
$\mathscr{L}:=\bigcup_{0<p\leq 1}\mathscr{L}_{p}$).

The weights we consider here are powers of the following function

(1.3)
$\mathcal{T}_{\Phi}^{\vec{\varepsilon}}(t)=1+\varepsilon_{1}\ln_{+}\left(\Phi\left(\frac{1}{t}\right)\right)+\varepsilon_{2}\ln_{+}\left(\Phi(t)\right),t>0.$

Here $\vec{\varepsilon}=(\varepsilon_{1},\varepsilon_{2})\in\\{0,1\\}^{2}$,
and $\Phi$ is a growth function.

In fact in the upper half plane, these weights are taken as powers of

$\mathcal{T}_{\Phi}^{\vec{\varepsilon}}(z)=1+\varepsilon_{1}\ln_{+}\left(\Phi\left(\frac{1}{\Im
mz}\right)\right)+\varepsilon_{2}\ln_{+}\left(\Phi(|z|)\right),(\varepsilon_{1},\varepsilon_{2})\in\\{0,1\\}^{2}.$

When $\Phi(t)=t$, and $(\varepsilon_{1},\varepsilon_{2})=(1,1)$, the
corresponding weight $\omega(z)$ is exactly the one that appears in the study
of Bloch functions of the upper half plane (see [6]).

Recall that a function $f$ is a Bloch function if

(1.4) $\Im m(z)|f^{\prime}(z)|\leq C.$

Let us put

$\|f\|_{*}:=|f(i)|+\sup_{z\in{\mathbb{C}}_{+}}\Im m(z)|f^{\prime}(z)|.$

Then one has (see [6]) that there is a constant $C>0$ such that for any Bloch
function $f$,

$|f(z)|\leq C\|f\|_{*}\left(\ln\left((e+\frac{1}{\Im
mz}\right)+\ln\left(e+|z|\right)\right),z\in{\mathbb{C}}_{+}.$

Our main result is as follows.

######  Theorem 1.

Let $1\leq p,q<\infty$, $\alpha,\beta>-1$. Assume that
$\Phi\in\mathscr{L}\cup\mathscr{U}$. Put
$\omega=\mathcal{T}_{\Phi}^{\vec{\varepsilon}}$, and let $k\in\mathbb{R}$.
Then the following assertions are equivalent.

  1. (a)

The operator $P_{\beta}$ is bounded on
$L^{p,q}({\mathbb{C}}_{+},\omega^{k}(\Im mz)dV_{\alpha}(z))$.

  2. (b)

The operator $P_{\beta}^{+}$ is bounded on
$L^{p,q}({\mathbb{C}}_{+},\omega^{k}(\Im mz)dV_{\alpha}(z))$.

  3. (c)

The parameters satisfy

(1.5) $\alpha+1<q(\beta+1).$

.

It is known that the boundedness of $P_{\beta}^{+}$ on
$L_{\alpha}^{q}(\mathbb{C}_{+})$ is equivalent to the boundedness of a
corresponding Hilbert-type operator of the positive line (see for example [1,
10]). The same observation can be made here. The Hilbert-type operator
$H_{\beta}$ is defined by

$H_{\beta}f(x)=\int_{0}^{\infty}\frac{f(y)y^{\beta}dy}{\left(x+y\right)^{1+\beta}}.$

Let $\Omega$ be a weight on $(0,\infty)$, and $1\leq p<\infty$. We denote by
$L_{\Omega}^{p}((0,\infty))$ the set of all measurable functions $f$ such that

$\|f\|_{\Omega,p}^{p}:=\int_{0}^{\infty}|f(x)|^{p}\Omega(x)dx<\infty.$

In particular, when $\Omega(x)=x^{\alpha}$, $\alpha\in\mathbb{R}$, we use the
notations $L_{\alpha}^{p}((0,\infty))=L^{p}((0,\infty),x^{\alpha}dx)$ for the
corresponding space and $\|\cdot\|_{\alpha,p}$ for the corresponding norm.

It is a well known that the operator $H_{\beta}$ is bounded on
$L_{\alpha}^{p}((0,\infty))$ for any $1<p<\infty$ if and only if the relation
$1+\alpha<p(1+\beta)$ holds (see for example [1]). We extend this result to
the weighted space $L^{p}((0,\infty),\omega(x)x^{\alpha}dx)$ for the above
family of weights. Our main results on the weighted boundedness of $H_{\beta}$
are given in the following.

######  Theorem 2.

Let $1\leq p<\infty$, $\alpha,\beta>-1$. Assume that
$\Phi\in\mathscr{L}\cup\mathscr{U}$. Put
$\omega=\mathcal{T}_{\Phi}^{\vec{\varepsilon}}$, and let $k\in\mathbb{R}$.
Then the following assertions are equivalent.

  1. (a)

The operator $H_{\beta}$ is bounded on
$L^{p}((0,\infty),\omega^{k}(y)y^{\alpha}dy)$.

  2. (b)

The parameters satisfy

(1.6) $\alpha+1<p(\beta+1).$

To prove the above result, we will prove a kind of weighted Forelli-Rudin type
estimate adapted to our setting and then combine it with a Schur’s lemma in
the case $p>1$.

######  Remark 1.

In general, to study the boundededness of operators with kernel on weighted
Lebesgue spaces, one needs to develop a weighted theory as the ones of
Muckenhoupt weights for Calderón-Zygmund operators (see for example [7]) or of
the Békollé-Bonami weights for the Bergman operator (see [2, 3]).Our family of
weights allows us to avoid this approach here and to reduce the question to a
relation between the parameters $\alpha,\beta$ and the exponent $p$. The
weighted theory for the operator $H_{\beta}$ (for general weights) will be
considered elsewhere.

The choice of the weights in this paper was inspired by [6] which also serves
as motivation for this paper. Among our applications, we have an atomic
decomposition of the above Bergman spaces which generalizes the work of J.
Gonessa on the same topic in [8].

The paper is organized as follows. In Section 2, we prove the result on the
Hilbert-type operators. The boundedness of the Bergman projection on mixed
norm spaces is proved in Section 3. In the last section, we provide some
applications of our results.

As usual, we use the the notation $A\lesssim B$ (respectively $A\gtrsim B$)
whenever there exists a uniform constant $C>0$ such that $A\leq CB$
(respectively $A\geq CB$). The notation $A\simeq B$ means that $A\lesssim B$
and $A\gtrsim B$.

##  2\. Boundedness of the Hilbert-type operator $H_{\beta}$

In this section, we consider the boundedness of the operator $H_{\beta}$ on
$L^{p}((0,\infty),\omega^{k}(y)y^{\alpha}dy)$.

###  2.1. Weighted estimates on the positive line

In the following, we give estimates of the measure of any interval $(0,t)$
with respect to $\omega^{k}(y)y^{\beta}dy$.

######  Lemma 1.

Let $\beta>-1$. Suppose that $\Phi\in\mathscr{L}\cup\mathscr{U}$. Put
$\omega=\mathcal{T}_{\Phi}^{\vec{\varepsilon}}$, and let $k>0$. Then the
following holds.

(2.1)
$\int_{0}^{t}\omega^{-k}(y)y^{\beta}dy\simeq\omega^{-k}(t)\,t^{1+\beta}.$

###### Proof.

We restrict ourselves to the case
$\vec{\varepsilon}=(\varepsilon_{1},\varepsilon_{2})=(1,1)$ as the other cases
follow the same way. Recall that in this case,

$\omega(y)=1+\ln_{+}\left(\Phi\left(\frac{1}{y}\right)\right)+\ln_{+}\left(\Phi(y)\right),$

and this is comparable to

$\ln\left(e+\Phi\left(\frac{1}{y}\right)\right)+\ln\left(e+\Phi(y)\right).$

We start by proving the upper estimate. If $0<t\leq 1$, then the estimate
above in (2.1) follows directly from the fact that $\omega$ is nonincreasing
in this interval.

More generally, let $t>0$. Then using that $\Phi$ satisfies (LABEL:uppertype)
or (LABEL:lowertype), and the notation

(2.2)
$p_{\Phi}=\left\\{\begin{array}[]{lcr}q&\mbox{if}&\Phi\in\mathscr{U}^{q}\\\
1&\mbox{if}&\Phi\in\mathscr{L}_{p}.\end{array},\right.$

we obtain

$\displaystyle I$ $\displaystyle:=$
$\displaystyle\int_{0}^{t}\omega^{-k}(y)y^{\beta}dy$ $\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}\int_{2^{-n-1}t}^{2^{-n}t}\frac{y^{\beta}dy}{\left[\ln\left(e+\Phi\left(\frac{1}{y}\right)\right)+\ln\left(e+\Phi(y)\right)\right]^{k}}$
$\displaystyle\leq$
$\displaystyle\frac{(2^{1+\beta}-1)}{1+\beta}t^{1+\beta}\sum_{n=0}^{\infty}\frac{2^{-(n+1)(1+\beta)}}{\left[\ln\left(e+\Phi\left(\frac{2^{n}}{t}\right)\right)+\ln\left(e+\Phi\left(\frac{t}{2^{n+1}}\right)\right)\right]^{k}}$
$\displaystyle\lesssim$ $\displaystyle
t^{1+\beta}\sum_{n=0}^{\infty}\frac{2^{-(n+1)(1+\beta)}\ln^{k}\left(2^{(n+1)p_{\Phi}}\right)}{\left[\ln\left(e+\Phi\left(\frac{1}{t}\right)\right)+\left[\ln\left(2^{(n+1)p_{\Phi}}\right)+\ln\left(e+\Phi\left(\frac{t}{2^{n+1}}\right)\right)\right]\right]^{k}}$
$\displaystyle\lesssim$ $\displaystyle\frac{t^{1+\beta}\left(p_{\Phi}\ln
2\right)^{k}}{\omega^{k}(t)}\sum_{n=0}^{\infty}2^{-(n+1)(1+\beta)}(n+1)^{k}$
$\displaystyle\lesssim$ $\displaystyle\frac{t^{1+\beta}}{\omega^{k}(t)}.$

The lower inequality follows by considering the integral over
$(\frac{t}{2},t)$. The proof is complete. ∎

We also have the following.

######  Lemma 2.

Let $\beta>-1$. Suppose that $\Phi\in\mathscr{L}\cup\mathscr{U}$. Put
$\omega=\mathcal{T}_{\Phi}^{\vec{\varepsilon}}$, and let $k>0$. Then the
following

(2.3) $\int_{0}^{t}\omega^{k}(y)y^{\beta}dy\simeq\omega^{k}(t)\,t^{1+\beta}.$

###### Proof.

The lower estimate is obtained as above. For the upper estimate, we again
restrict to the weight

$\omega(y)=\ln\left(e+\Phi\left(\frac{1}{y}\right)\right)+\ln\left(e+\Phi(y)\right).$

If $t\geq 1$, then the upper estimate in (2.3) follows directly from the fact
that $\omega$ is nonincreasing in this range.

More generally, let $t>0$. We obtain

$\displaystyle I$ $\displaystyle:=$
$\displaystyle\int_{0}^{t}\omega^{k}(y)y^{\beta}dy$ $\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}\int_{2^{-n-1}t}^{2^{-n}t}\left[\ln\left(e+\Phi\left(\frac{1}{y}\right)\right)+\ln\left(e+\Phi(y)\right)\right]^{k}y^{\beta}dy$
$\displaystyle\lesssim$
$\displaystyle(2^{1+\beta}-1)t^{1+\beta}\sum_{n=0}^{\infty}2^{-(n+1)(1+\beta)}\left[\ln\left(e+\Phi\left(\frac{2^{n+1}}{t}\right)\right)+\ln\left(e+\Phi(2^{-n}t)\right)\right]^{k}$
$\displaystyle\lesssim$ $\displaystyle
t^{1+\beta}\sum_{n=0}^{\infty}2^{-(n+1)(1+\beta)}\left[(n+1)p_{\Phi}\ln
2+\ln\left(e+\Phi\left(\frac{1}{t}\right)\right)+\ln\left(e+\Phi(t)\right)\right]^{k}$
$\displaystyle\leq$ $\displaystyle
t^{1+\beta}\left(p_{\Phi}\right)^{k}\omega^{k}(t)\sum_{n=0}^{\infty}2^{-(n+1)(1+\beta)}(n+1)^{k}$
$\displaystyle\lesssim$ $\displaystyle t^{1+\beta}\omega^{k}(t).$

The proof is complete. ∎

We have the following estimate.

######  Lemma 3.

Let $\alpha>0$, and $\beta>-1$. Suppose that
$\Phi\in\mathscr{L}\cup\mathscr{U}$. Put
$\omega=\mathcal{T}_{\Phi}^{\vec{\varepsilon}}$, and let $k>0$. Then the
following holds

(2.4)
$\int_{0}^{\infty}\frac{\omega^{-k}(y)y^{\beta}dy}{\left(x+y\right)^{1+\alpha+\beta}}\simeq\omega^{-k}(x)\,x^{-\alpha},x>0.$

###### Proof.

Again we only present the proof for $\vec{\varepsilon}=(1,1)$. For $x>0$
fixed, let $I_{0}=(0,x]$. For $j\in\mathbb{N}$, let $I_{j}=(0,2^{j}x]$. Then
define $J_{0}=I_{0}$, and for $j\geq 1$, $J_{j}=I_{j}\setminus I_{j-1}$.

We obtain using Lemma 1 that

$\displaystyle L$ $\displaystyle:=$
$\displaystyle\int_{0}^{\infty}\frac{\omega^{-k}(y)y^{\beta}dy}{\left(x+y\right)^{1+\alpha+\beta}}$
$\displaystyle=$
$\displaystyle\sum_{j=0}^{\infty}\int_{J_{j}}\frac{\omega^{-k}(y)y^{\beta}dy}{\left(x+y\right)^{1+\alpha+\beta}}$
$\displaystyle\leq$ $\displaystyle
x^{-(1+\alpha+\beta)}\sum_{j=0}^{\infty}2^{-(j-1)(1+\alpha+\beta)}\int_{I_{j}}\omega^{-k}(y)y^{\beta}dy$
$\displaystyle\lesssim$ $\displaystyle
x^{-(1+\alpha+\beta)}\sum_{j=0}^{\infty}2^{-(j-1)(1+\alpha+\beta)}\frac{\left(2^{j}x\right)^{1+\beta}}{\omega^{k}(2^{j}x)}.$

If $x\geq 1$, then we directly have that $\omega(2^{j}x)>\omega(x)$. If
$0<x<1$, then for any $j>0,$ we obtain

$\displaystyle\omega(2^{j}x)$ $\displaystyle=$
$\displaystyle\ln\left(e+\Phi\left(2^{j}x\right)\right)+\ln\left(e+\Phi\left(\frac{1}{2^{j}x}\right)\right)$
$\displaystyle\gtrsim$
$\displaystyle\frac{\ln\left(e+\Phi(x)\right)+\ln\left(2^{jp_{\Phi}}\right)+\ln\left(e+\Phi\left(\frac{1}{2^{j}x}\right)\right)}{\ln\left(2^{jp_{\Phi}}\right)}$
$\displaystyle\gtrsim$
$\displaystyle\frac{\ln\left(e+\Phi(x)\right)+\ln\left(e+\Phi\left(\frac{1}{x}\right)\right)}{\ln\left(2^{jp_{\Phi}}\right)}.$

Hence we finally obtain

$\displaystyle L$ $\displaystyle\lesssim$
$\displaystyle\omega^{-k}(x)\,x^{-\alpha}\sum_{j=0}^{\infty}2^{-(j-1)\alpha}j^{k}$
$\displaystyle\lesssim$ $\displaystyle\omega^{-k}(x)\,x^{-\alpha}.$

The lower estimate is obtained by taking the integral from $\frac{x}{2}$ to
$x$. The proof is complete. ∎

We have the following.

######  Lemma 4.

Let $\alpha>0$, and $\beta>-1$. Suppose that
$\Phi\in\mathscr{L}\cup\mathscr{U}$. Put
$\omega=\mathcal{T}_{\Phi}^{\vec{\varepsilon}}$, and let $k>0$. Then the
following holds

(2.5)
$\int_{0}^{\infty}\frac{\omega^{k}(y)y^{\beta}dy}{\left(x+y\right)^{1+\alpha+\beta}}\simeq\omega^{k}(x)\,x^{-\alpha},x>0.$

###### Proof.

The proof is essentially the same as above. One only needs the upper estimate
of $\omega(2^{j}x)$, $j>0$. If $x>0$, then

$\displaystyle\omega(2^{j}x)$ $\displaystyle\lesssim$
$\displaystyle\ln\left(e+\Phi\left(\frac{1}{x}\right)\right)+\ln\left(2^{jp_{\Phi}}\right)+\ln\left(e+\Phi(x)\right)$
$\displaystyle=$
$\displaystyle\ln\left(e+\Phi\left(\frac{1}{x}\right)\right)+jp_{\Phi}\ln
2+\ln\left(e+\Phi(x)\right)$ $\displaystyle\lesssim$ $\displaystyle
j\omega(x).$

The estimates then follow as in the previous lemma. ∎

###  2.2. Proofs of Theorem 2

We start by considering the case $p=1$. We prove the following.

######  Theorem 3.

Let $\alpha,\beta>-1$. Assume that $\Phi\in\mathscr{L}\cup\mathscr{U}$. Put
$\omega=\mathcal{T}_{\Phi}^{\vec{\varepsilon}}$, and let $k\in\mathbb{R}$.
Then the following assertions are equivalent.

  1. (a)

The operator $H_{\beta}$ is bounded on
$L^{1}((0,\infty),\omega^{k}(y)y^{\alpha}dy)$.

  2. (b)

The parameters satisfy

$\beta>\alpha.$

###### Proof.

That (b) implies (a) follows from Fubini’s theorem and Lemma 3 and Lemma 4.

Conversely, suppose that $H_{\beta}$ is bounded on
$L^{1}((0,\infty),\omega^{-k}(y)y^{\alpha}dy)$, $k>0$. Then its adjoint
$H_{\beta}^{*}$ defined by

$H_{\beta}^{*}f(x)=\omega^{k}(x)x^{\beta-\alpha}\int_{0}^{\infty}\frac{f(y)\omega^{-k}(y)y^{\alpha}dy}{\left(x+y\right)^{1+\beta}}$

is bounded on $L^{\infty}((0,\infty))$. Take $f(y)=\chi_{[1,2]}(y)$. Then

$H_{\beta}^{*}f(x)\simeq\frac{\omega^{k}(x)x^{\beta-\alpha}}{\left(x+1\right)^{1+\beta}}.$

Clearly, for $0<x<1$, one has that $H_{\beta}^{*}f(x)\gtrsim
x^{\beta-\alpha}.$ Hence the boundedness of $H_{\beta}^{*}f(x)$ implies that
$x^{\beta-\alpha}$ is bounded on $(0,1)$. Thus, necessarily, $\beta-\alpha>0$.

Now, suppose that $H_{\beta}$ is bounded on
$L^{1}((0,\infty),\omega^{k}(y)y^{\alpha}dy)$, $k>0$ i.e there exists a
constant $C>0$ such that for any $f\in
L^{1}((0,\infty),\omega^{k}(y)y^{\alpha}dy)$,

(2.6) $\int_{0}^{\infty}|H_{\beta}f(x)|\omega^{k}(x)x^{\alpha}dx\leq
C\int_{0}^{\infty}|f(x)|\omega^{k}(x)x^{\alpha}dx$

Take $f(y)=\chi_{[1,2]}(y)$. Then

$H_{\beta}f(x)\simeq\frac{1}{(x+1)^{1+\beta}}$

and

$\int_{0}^{\infty}|f(x)|\omega^{k}(x)x^{\alpha}dx\simeq 1.$

Taking this into (2.6), we obtain

$\int_{0}^{\infty}\frac{x^{\alpha}}{(x+1)^{1+\beta}}dx\leq\int_{0}^{\infty}\frac{\omega^{k}(x)x^{\alpha}}{(x+1)^{1+\beta}}dx<\infty.$

This necessarily implies that $1+\beta>\alpha+1$. The proof is complete. ∎

We next consider the case $p>1$.

######  Theorem 4.

Let $1<p<\infty$, and $\alpha,\beta>-1$. Assume that
$\Phi\in\mathscr{L}\cup\mathscr{U}$. Put
$\omega=\mathcal{T}_{\Phi}^{\vec{\varepsilon}}$, and let $k\in\mathbb{R}$.
Then the following assertions are equivalent.

  1. (a)

The operator $H_{\beta}$ is bounded on
$L^{p}((0,\infty),\omega^{k}(y)y^{\alpha}dy)$.

  2. (b)

The parameters satisfy (1.6).

###### Proof.

We restrict to the case of $\omega^{-k}$, $k>0$ as the other case follows from
the same ideas. Recall that

$H_{\beta}f(x)=\int_{0}^{\infty}\frac{f(y)y^{\beta}dy}{\left(x+y\right)^{1+\beta}}.$

We note that the kernel of $H_{\beta}$ with respect to the measure
$\omega^{-k}(y)y^{\alpha}dy$ is

$K(x,y)=\frac{\omega^{k}(y)y^{\beta-\alpha}}{\left(x+y\right)^{1+\beta}}.$

Assume that (1.6) holds. By the Schur’s lemma (see [9]), to prove that
$H_{\beta}$ is bounded on $L^{p}((0,\infty),\omega^{-k}(y)y^{\alpha}dy)$, it
suffices to find a positive function $\phi$ such that

(2.7)
$\int_{0}^{\infty}K(x,y)\phi^{p^{\prime}}(y)\omega^{-k}(y)y^{\alpha}dy\leq
C\phi^{p^{\prime}}(x)$

and

(2.8) $\int_{0}^{\infty}K(x,y)\phi^{p}(x)\omega^{-k}(x)x^{\alpha}dx\leq
C\phi^{p}(y).$

Let us take $\phi(t)=t^{-\frac{\alpha+1}{pp^{\prime}}}$. Then using Lemma 3,
we obtain

$\displaystyle L$ $\displaystyle:=$
$\displaystyle\int_{0}^{\infty}K(x,y)\phi^{p^{\prime}}(y)\omega^{-k}(y)y^{\alpha}dy$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}\frac{y^{\beta-\frac{\alpha+1}{p}}dy}{\left(x+y\right)^{1+\beta}}$
$\displaystyle\simeq$ $\displaystyle
x^{-\frac{\alpha+1}{p}}=\phi^{p^{\prime}}(x).$

In the same way, we have

$\displaystyle L$ $\displaystyle:=$
$\displaystyle\int_{0}^{\infty}K(x,y)\phi^{p}(x)\omega^{-k}(x)x^{\alpha}dx$
$\displaystyle=$
$\displaystyle\omega^{k}(y)y^{\beta-\alpha}\int_{0}^{\infty}\frac{\omega^{-k}(x)x^{-\frac{\alpha+1}{p^{\prime}}+\alpha}dx}{\left(x+y\right)^{1+\beta}}$
$\displaystyle\simeq$ $\displaystyle
x^{-\frac{\alpha+1}{p^{\prime}}}=\phi^{p}(x).$

This completes the proof of the sufficient part.

Now suppose that $H_{\beta}$ is bounded on
$L^{p}((0,\infty),\omega^{-k}(y)y^{\alpha}dy)$. Then its adjoint
$H_{\beta}^{*}$ defined by

$H_{\beta}^{*}f(x)=\omega^{k}(x)x^{\beta-\alpha}\int_{0}^{\infty}\frac{f(y)\omega^{-1}(y)y^{\alpha}dy}{\left(x+y\right)^{1+\beta}}$

is bounded on $L^{p^{\prime}}((0,\infty),\omega^{-k}(y)y^{\alpha}dy)$. Take
$f(y)=\chi_{[1,2]}(y)$. Then

$H_{\beta}^{*}f(x)\simeq\frac{\omega^{k}(x)x^{\beta-\alpha}}{\left(x+1\right)^{1+\beta}}\geq\frac{\omega^{\frac{k}{p^{\prime}}}(x)x^{\beta-\alpha}}{\left(x+1\right)^{1+\beta}}.$

Thus that

$\|H_{\beta}^{*}f\|_{L_{\omega^{-k},\alpha}^{p^{\prime}}}\leq
C\|f\|_{L_{\omega^{-k},\alpha}^{p^{\prime}}}\simeq C$

implies that

$\int_{0}^{\infty}\frac{x^{(\beta-\alpha)p^{\prime}+\alpha}}{\left(x+1\right)^{(1+\beta)p^{\prime}}}dx<\infty.$

Thus, necessarily, $(\beta-\alpha)p^{\prime}+\alpha>-1$. That is (1.6) holds.
The proof is complete. ∎

######  Remark 2.

To prove the implication (a)$\Rightarrow$(b) in the case of $\omega^{k}$,
$k>0$ in the above result, one can work directly with $H_{\beta}$ and test the
boundedness on the function $f(y)=\chi_{[1,2]}(y)$ as in Theorem 3.

##  3\. Boundedness of the Bergman projection

###  3.1. Proofs of Theorem 1

We prove the various cases appearing in Theorem 1.

######  Proposition 1.

Let $1<p,q<\infty$, $\alpha,\beta>-1$. Assume that
$\Phi\in\mathscr{L}\cup\mathscr{U}$. Put
$\omega=\mathcal{T}_{\Phi}^{\vec{\varepsilon}}$, and let $k\in\mathbb{R}$.
Then the following assertions are equivalent.

  1. (a)

The operator $P_{\beta}$ is bounded on
$L^{p,q}({\mathbb{C}}_{+},\omega^{k}dV_{\alpha})$.

  2. (b)

The operator $P_{\beta}^{+}$ is bounded on
$L^{p,q}({\mathbb{C}}_{+},\omega^{k}dV_{\alpha})$.

  3. (c)

The parameters satisfy

$\alpha+1<q(\beta+1).$

.

###### Proof.

The implication (b)$\Rightarrow$(a) is clear. Let us prove that
(c)$\Rightarrow$(b) and (a)$\Rightarrow$(c).

(c)$\Rightarrow$(b): Let write $f_{y}(x)=f(x+iy)$. Then it is easy to see (see
for example [1, Lemma 5.1]) that

$\left\|\left(P_{\beta}^{+}f\right)_{y}\right\|_{L^{p}(\mathbb{R})}\leq
C_{\beta}H_{\beta}(\|f_{v}\|_{L^{p}(\mathbb{R})})(y).$

Thus the boundedness of $H_{\beta}$ on
$L^{q}((0,\infty)\omega^{k})(y)y^{\alpha}dy$ implies the boundedness of
$P_{\beta}^{+}$ on $L^{p,q}(\mathbb{C}_{+},\omega^{k}dV_{\alpha})$. Hence the
implication follows from this observation and Theorem 2.

(a)$\Rightarrow$(c): Assume that $P_{\beta}$ is bounded on
$L^{p,q}({\mathbb{C}}_{+},\omega^{-k}(y)y^{\alpha}dxdy)$, $k>0$. Then its
adjoint $P_{\beta}^{*}$ defined by

$P_{\beta}^{*}f(z)=\omega^{k}(\Im mz)\left(\Im
mz\right)^{\beta-\alpha}\int_{\mathbb{C}_{+}}\frac{f(w)}{\left(z-\bar{w}\right)^{2+\beta}}\omega^{-k}(\Im
m\,w)dV_{\alpha}(w)$

is bounded on
$L^{p^{\prime},q^{\prime}}({\mathbb{C}}_{+},\omega^{-k}(y)y^{\alpha}dxdy)$.
Let us take $f(w)=\omega^{k}(\Im m\,w)\left(\Im
m\,w\right)^{-\alpha}\chi_{B(i,1)}(w)$ where $B(i,1)$ is the euclidean ball
centered at $i$ with radius $1$. Applying the mean value property, we obtain
that

$P_{\beta}^{*}f(z)=c\frac{\omega^{k}(\Im mz)\left(\Im
mz\right)^{\beta-\alpha}}{\left(z+i\right)^{2+\beta}}.$

Thus as $\omega(y)\geq 1$, we obtain that

$\|P_{\beta}^{*}f\|_{L^{p^{\prime},q^{\prime}}(\omega^{-k}dV_{\alpha})}\leq
C\|f\|_{L^{p^{\prime},q^{\prime}}(\omega^{-k}dV_{\alpha})}=C$

implies that the function $z\longrightarrow\frac{\left(\Im
mz\right)^{\beta-\alpha}}{\left(z+i\right)^{2+\beta}}$ belongs to
$L^{p^{\prime},q^{\prime}}({\mathbb{C}}_{+},dV_{\alpha})$. In particular, we
have that

$\int_{0}^{1}y^{(\beta-\alpha)q^{\prime}+\alpha}dy\lesssim\int_{0}^{1}\left(\int_{-\frac{1}{4}}^{\frac{1}{4}}\frac{dx}{|x+i(y+1)|^{(2+\beta)p^{\prime}}}\right)^{\frac{q^{\prime}}{p^{\prime}}}y^{(\beta-\alpha)q^{\prime}+\alpha}dy<\infty.$

Hence one necessarily has that $(\beta-\alpha)q^{\prime}+\alpha>-1$ or
equivalently, $\alpha+1<q(\beta+1)$.

Next, suppose that $P_{\beta}$ is bounded on
$L^{p,q}({\mathbb{C}}_{+},\omega^{k}(y)y^{\alpha}dxdy)$, $k>0$. Then its
adjoint $P_{\beta}^{*}$ defined by

$P_{\beta}^{*}f(z)=\omega^{-k}(\Im mz)\left(\Im
mz\right)^{\beta-\alpha}\int_{\mathbb{C}_{+}}\frac{f(w)}{\left(z-\bar{w}\right)^{2+\beta}}\omega^{k}(\Im
m\,w)dV_{\alpha}(w)$

is bounded on
$L^{p^{\prime},q^{\prime}}({\mathbb{C}}_{+},\omega^{k}(y)y^{\alpha}dxdy)$. Let
us take $f(w)=\omega^{-k}(\Im m\,w)\left(\Im
m\,w\right)^{-\alpha}\chi_{B(i,1)}(w)$ where $B(i,1)$ is again the euclidean
ball centered at $i$ with radius $1$. Applying the mean value property again,
we obtain that

$P_{\beta}^{*}f(z)=c\frac{\omega^{-k}(\Im mz)\left(\Im
mz\right)^{\beta-\alpha}}{\left(z+i\right)^{2+\beta}}.$

Thus

$\|P_{\beta}^{*}f\|_{L^{p^{\prime},q^{\prime}}(\omega^{k}dV_{\alpha})}\leq
C\|f\|_{L^{p^{\prime},q^{\prime}}(\omega^{k}dV_{\alpha})}=C$

implies that

(3.1)
$\int_{0}^{1}\omega^{-kq^{\prime}}(y)y^{(\beta-\alpha)q^{\prime}+\alpha}dy\lesssim\int_{0}^{1}\left(\int_{-\frac{1}{4}}^{\frac{1}{4}}\frac{dx}{|x+i(y+1)|^{(2+\beta)p^{\prime}}}\right)^{\frac{q^{\prime}}{p^{\prime}}}\omega^{-kq^{\prime}+k}(y)y^{(\beta-\alpha)q^{\prime}+\alpha}dy<\infty.$

To prove that necessarily, $(\beta-\alpha)q^{\prime}+\alpha+1>0$, it suffices
to show that the left hand side integral in (3.1) is not convergent if
$(\beta-\alpha)q^{\prime}+\alpha+1\leq 0$. In fact it suffices to prove this
for $(\beta-\alpha)q^{\prime}+\alpha+1=0$. Suppose that
$(\beta-\alpha)q^{\prime}+\alpha+1=0$. From (3.1) we have that

$\displaystyle\infty$ $\displaystyle>$
$\displaystyle\int_{0}^{1}\omega^{-kq^{\prime}}(y)y^{-1}dy$
$\displaystyle\gtrsim$
$\displaystyle\int_{0}^{1}\left(\ln\Phi\left(\frac{e}{y}\right)\right)^{-kq^{\prime}}y^{-1}dy$
$\displaystyle\gtrsim$
$\displaystyle\int_{0}^{1}\left(\frac{e}{y}\right)^{-p_{\Phi}kq^{\prime}}y^{-1}dy.$

This is impossible since $\int_{0}^{1}y^{-1-p_{\Phi}kq^{\prime}}dy$ is not
convergent. The proof is complete. ∎

In the case $q=1$, we have the following.

######  Proposition 2.

Let $1\leq p<\infty$, $\alpha,\beta>-1$. Assume that
$\Phi\in\mathscr{L}\cup\mathscr{U}$. Put
$\omega=\mathcal{T}_{\Phi}^{\vec{\varepsilon}}$, and let $k\in\mathbb{R}$.
Then the following assertions are equivalent.

  1. (a)

The operator $P_{\beta}$ is bounded on
$L^{p,1}({\mathbb{C}}_{+},\omega^{k}dV_{\alpha})$.

  2. (b)

The operator $P_{\beta}^{+}$ is bounded on
$L^{p,1}({\mathbb{C}}_{+},\omega^{k}dV_{\alpha})$.

  3. (c)

The parameters satisfy

$\alpha<\beta.$

###### Proof.

The implications (b)$\Rightarrow$(a) and (c)$\Rightarrow$(b) follow as above
with the help of Theorem 3.

Reasoning as above, we obtain that the functions
$z\longrightarrow\frac{\left(\Im
mz\right)^{\beta-\alpha}}{\left(z+i\right)^{2+\beta}}$ and
$z\longrightarrow\frac{\omega^{-k}(\Im mz)\left(\Im
mz\right)^{\beta-\alpha}}{\left(z+i\right)^{2+\beta}}$ belong to
$L^{p^{\prime},\infty}(\mathbb{C}_{+})$. This necessarily leads to
$\beta>\alpha$. ∎

##  4\. Applications

We provide in this section, three applications of our above results. For our
purpose in this section, we need to recall the definition of the Hardy spaces.
For $1\leq p<\infty$, the Hardy space $H^{p}(\mathbb{C}_{+})$ consists of all
holomorphic functions $f$ on $\mathbb{C}_{+}$ such that

$\|f\|_{p}:=\sup_{y>0}\left(\int_{\mathbb{R}}|f(x+iy)|^{p}dx\right)^{1/p}<\infty.$

###  4.1. Reproducing formulas and duality

We prove reproducing formula only in the case of the weight $\omega^{-k}$ with
$k>0$ as the corresponding spaces are larger than the usual mixed norm Bergman
spaces that correspond to the case $\omega=1$ for which reproducing formulas
are known (see for example [4, 5, 8]).

Put $\Phi\in\mathscr{L}\cup\mathscr{U}$. Put
$\omega=\mathcal{T}_{\Phi}^{\vec{\varepsilon}}$. For $f\in
A^{p,q}_{\omega^{-k},\alpha}(\mathbb{C}_{+})$, $k>0$, and for $x\in\mathbb{R}$
and $y>0$ given, defined $f_{1}$ and $f_{2}$ by

$f_{1}(u+iv)=f(x+u+iv),\text{and}\quad f_{2}(u+iv)=f(y(u+iv)).$

Then

$\|f_{1}\|_{A^{p,q}_{\omega^{-k},\alpha}}=\|f\|_{A^{p,q}_{\omega^{-k},\alpha}}$

and

$\|f_{2}\|_{A^{p,q}_{\omega^{-k},\alpha}}=y^{-\frac{1+\alpha}{q}-\frac{1}{p}}\|f\|_{A^{p,q}_{\omega_{y^{-1}}^{-k},\alpha}}$

where $\omega_{y^{-1}}(v)=\omega(y^{-1}v)$. Moreover, we have the estimate

(4.1)
$\omega^{-1}(y^{-1}v)\lesssim\left(1+\epsilon_{1}\ln_{+}\left(\frac{1}{y}\right)+\epsilon_{2}\ln_{+}(y)\right)\omega^{-1}(v).$

For simplicity, we put

$\omega_{0}(y)=1+\epsilon_{1}\ln_{+}\left(\frac{1}{y}\right)+\epsilon_{2}\ln_{+}(y).$

We start this part by the following result which follows as in [8, Proposition
1.3.4] (see also [5]) with the help of the above observations.

######  Proposition 3.

  * Let $1\leq p,q<\infty$, and $\alpha>-1$. Assume that $\Phi\in\mathscr{L}\cup\mathscr{U}$. Put $\omega=\mathcal{T}_{\Phi}^{\vec{\varepsilon}}$, and let $k>0$. Then the following assertions are satisfied.

  * (i)

There exists a constant $C>0$ such that for all $x+iy\in\mathbb{C}_{+}$ and
all $f\in A^{p,q}_{\omega^{-k},\alpha}(\mathbb{C}_{+})$, the following
inequality holds.

(4.2) $|f(x+iy)|\leq
C\left(\omega_{0}(y)\right)^{\frac{k}{q}}y^{-\frac{1+\alpha}{q}-\frac{1}{p}}\|f\|_{A^{p,q}_{\omega^{-k},\alpha}}.$

  * (ii)

There exists a constant $C>0$ such that for all $y\in(0,\infty)$ and all $f\in
A^{p,q}_{\omega^{-k},\alpha}(\mathbb{C}_{+})$,

(4.3) $\|f_{y}\|_{p}:=\|f(\cdot+iy)\|_{L^{p}(\mathbb{R})}\leq
C\left(\omega_{0}(y)\right)^{\frac{k}{q}}y^{-\frac{1+\alpha}{q}}\|f\|_{A^{p,q}_{\omega^{-k},\alpha}}.$

  * (iii)

For all $f\in A^{p,q}_{\omega^{-k},\alpha}(\mathbb{C}_{+})$, and for all
$y>0$, the following holds:

$\lim_{|x|\rightarrow\infty}|f(x+iy)|=0=\lim_{|x|+y\rightarrow\infty,y>y_{0}}|f(x+iy)|.$

We then obtain the following.

######  Proposition 4.

Let $1\leq p,q<\infty$, and $\alpha>-1$. Assume that
$\Phi\in\mathscr{L}\cup\mathscr{U}$. Put
$\omega=\mathcal{T}_{\Phi}^{\vec{\varepsilon}}$, and let $k>0$. Let $f\in
A^{p,q}_{\omega^{-k},\alpha}(\mathbb{C}_{+})$. Then the following assertions
are satisfied.

  * (i)

The function $y\mapsto\|f_{y}\|_{p}$ is non-increasing and continuous on
$(0,\infty)$.

  * (ii)

$f(\cdot+i\varepsilon)$ is in $A^{p,q}_{\omega^{-k},\alpha}(\mathbb{C}_{+})$
for any $\varepsilon>0$, and tends to $f$ in
$A^{p,q}_{\omega^{-k},\alpha}(\mathbb{C}_{+})$ as $\varepsilon$ tends to zero.

###### Proof.

Assertion (ii) in Proposition 3, means that $f_{y}$ is in
$H^{p}(\mathbb{C}_{+})$, hence (i) holds. We proceed to prove (ii): it follows
using (i) that

$\displaystyle\|f(\cdot+i\varepsilon)\|_{A^{p,q}_{\omega^{-k},\alpha}}^{q}$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}\|f_{y+\varepsilon}\|_{p}^{q}\,\omega^{-k}(y)y^{\alpha}dy$
$\displaystyle\leq$
$\displaystyle\int_{0}^{\infty}\|f_{y}\|_{p}^{q}\,\omega^{-k}(y)y^{\alpha}dy$
$\displaystyle=$ $\displaystyle\|f\|_{A^{p,q}_{\omega^{-k},\alpha}}^{q}.$

We have that $f(\cdot+i\varepsilon)$ has $f$ as limit as
$\varepsilon\rightarrow 0$. Hence by (i) and [11, Theorem 5.6], we have that

$\lim_{\varepsilon\rightarrow 0}\|f_{y+\varepsilon}-f_{y}\|_{p}=0$

It follows from the above and the Dominated Convergence Theorem that

$\lim_{\varepsilon\rightarrow
0}\|f(\cdot+i\varepsilon)-f\|_{A^{p,q}_{\omega^{-k},\alpha}}^{q}=\lim_{\varepsilon\rightarrow
0}\int_{0}^{\infty}\|f_{y+\varepsilon}-f_{y}\|_{p}^{q}\,\omega^{-k}(y)y^{\alpha}dy=0.$

∎

Let us prove the following.

######  Proposition 5.

Let $1\leq p,q<\infty$, and $\alpha>-1$. Assume that
$\Phi\in\mathscr{L}\cup\mathscr{U}$. Put
$\omega=\mathcal{T}_{\Phi}^{\vec{\varepsilon}}$, and let $k>0$. Then
$A^{p,q}_{\alpha}(\mathbb{C}_{+})$ is a dense subspace of
$A^{p,q}_{\omega^{-k},\alpha}(\mathbb{C}_{+})$.

###### Proof.

For $m>0$ a large enough integer, we have that for any $\varepsilon>0$, the
function $g_{\varepsilon}(z)=(1-i\varepsilon z)^{-m}$ is in
$A^{p,q}_{\alpha}(\mathbb{C}_{+})$. Let $f\in
A^{p,q}_{\omega^{-k},\alpha}(\mathbb{C}_{+})$, and define $F^{(\varepsilon)}$
by

$F^{(\varepsilon)}(z)=g_{\varepsilon}(z)f(z+i\varepsilon).$

Then by using assertion (ii) in Proposition 4, one sees that
$F^{(\varepsilon)}$ belongs to $A^{p,q}_{\omega^{-k},\alpha}(\mathbb{C}_{+})$.
Also if $\rho(t)=1+\frac{kp}{p(1+\alpha)+q}+\ln(e+t)$ (one can also use
$\ln(e+\frac{1}{t})$ when $t$ is small), then as the function
$y\mapsto(\rho(y))^{\frac{k}{q}}y^{-\frac{1+\alpha}{q}-\frac{1}{p}}$ is non-
increasing, one obtains from assertion (i) in Proposition 3 that the factor
$f(z+i\varepsilon)$ is bounded and hence that $F^{(\varepsilon)}$ belongs to
$A^{p,q}_{\alpha}(\mathbb{C}_{+})$.

We have that $F^{(\varepsilon)}\rightarrow f$ as $\varepsilon\rightarrow 0$.
The Dominated Convergence Theorem then leads to

$\lim_{\varepsilon\rightarrow
0}\|F^{(\varepsilon)}-f\|_{A^{p,q}_{\omega^{-k},\alpha}}=0.$

∎

We will need the following.

######  Proposition 6.

Let $1\leq p,q<\infty$, and $\alpha>-1$. Assume that
$\Phi\in\mathscr{L}\cup\mathscr{U}$. Put
$\omega=\mathcal{T}_{\Phi}^{\vec{\varepsilon}}$, and let $k>0$. Let $\beta>-1$
with $1+\alpha<q(1+\beta)$. Then the Bergman projection $P_{\beta}$ reproduces
the functions in $A^{p,q}_{\omega^{-k},\alpha}(\mathbb{C}_{+})$.

###### Proof.

As $1+\alpha<q(1+\beta)$, $P_{\beta}$ reproduces the elements of
$A^{p,q}_{\alpha}(\mathbb{C}_{+})$, Proposition 5 allows us to conclude that
this projector also reproduces functions in
$A^{p,q}_{\omega^{-k},\alpha}(\mathbb{C}_{+})$. ∎

It follows from classical arguments and the above reproducing formula that the
following holds.

######  Proposition 7.

Let $1<p,q<\infty$, and $\alpha>-1$. Assume that
$\Phi\in\mathscr{L}\cup\mathscr{U}$. Put
$\omega=\mathcal{T}_{\Phi}^{\vec{\varepsilon}}$, and let $k\in\mathbb{R}$.
Then the dual space
$\left(A^{p,q}_{\omega^{k},\alpha}(\mathbb{C}_{+})\right)^{*}$ of the Bergman
space $A^{p,q}_{\omega^{k},\alpha}(\mathbb{C}_{+})$ identifies with
$A^{p^{\prime},q^{\prime}}_{\omega^{k(1-q^{\prime})},\alpha}(\mathbb{C}_{+})$
under the duality pairing

$\langle
f,g\rangle_{\alpha}:=\int_{\mathbb{C}_{+}}f(z)\overline{g(z)}dV_{\alpha}(z),f\in
A^{p,q}_{\omega^{k},\alpha}(\mathbb{C}_{+}),g\in
A^{p^{\prime},q^{\prime}}_{\omega^{k(1-q^{\prime})},\alpha}(\mathbb{C}_{+}).$

###  4.2. Derivative characterization of weighted mixed norm Bergman spaces

In this section, we prove that $f\in
A^{p,q}_{\omega^{k},\alpha}(\mathbb{C}_{+})$ if and only if the function
$x+iy\mapsto yf^{\prime}(x+iy)$ belongs to
$L^{p,q}_{\omega^{k},\alpha}(\mathbb{C}_{+})$.

######  Theorem 5.

Let $1<p,q<\infty$, and $\alpha>-1$. Assume that
$\Phi\in\mathscr{L}\cup\mathscr{U}$. Put
$\omega=\mathcal{T}_{\Phi}^{\vec{\varepsilon}}$, and let $k\in\mathbb{R}$. Let
$f$ be a holomorphic function on $\mathbb{C}_{+}$. Then the following
assertions are equivalent.

  * (a)

$f$ belongs to $A^{p,q}_{\omega^{k},\alpha}(\mathbb{C}_{+})$.

  * (b)

The function $x+iy\mapsto yf^{\prime}(x+iy)$ belongs to
$L^{p,q}_{\omega^{k},\alpha}(\mathbb{C}_{+})$.

###### Proof.

(a)$\Rightarrow$(b): Recall the following inequality for the derivative
$f^{\prime}$ of the holomorphic function $f$ on the upper half-plane (see [5,
Problem 2])

$y|f^{\prime}(x+iy)|\leq\frac{C}{y^{2}}\int_{\frac{y}{2}<v<2y}\int_{|x-u|<y}|f(u+iv)|dudv.$

Using Minkowski’s inequality, we obtain

$y\|f^{\prime}(\cdot+iy)\|_{p}\leq\left(\frac{C}{y}\int_{\frac{y}{2}<v<2y}\|f(\cdot+iv)\|_{p}^{p}dv\right)^{\frac{1}{p}}.$

Since the function $v\mapsto\|f(\cdot+iv)\|_{p}$ is non-increasing, we deduce
that

$y\|f^{\prime}(\cdot+iy)\|_{p}\leq C\|f(\cdot+iy/2)\|_{p}.$

It follows that

$\displaystyle\int_{0}^{\infty}\|f^{\prime}(\cdot+iy)\|_{p}^{q}y^{q+\alpha}\omega^{k}(y)dy$
$\displaystyle\leq$ $\displaystyle
C\int_{0}^{\infty}\|f(\cdot+iy/2)\|_{p}^{q}y^{\alpha}\omega^{k}(y)dy$
$\displaystyle\lesssim$
$\displaystyle\int_{0}^{\infty}\|f(\cdot+iy)\|_{p}^{q}y^{\alpha}\omega^{k}(y)dy.$

(b)$\Rightarrow$(a): Using assertion (iii) of Proposition 3, one obtains that

$|f(x+iy)|=\left|\int_{y}^{\infty}-f^{\prime}(x+iv)dv\right|\leq\int_{y}^{\infty}|f^{\prime}(x+iv)|dv.$

Using the above and Minkowski’s inequality, we derive that

$\displaystyle\|f(\cdot+iy)\|_{p}$ $\displaystyle\leq$
$\displaystyle\left(\int_{\mathbb{R}}\left(\int_{y}^{\infty}|f^{\prime}(x+iv)|dv\right)^{p}dx\right)^{1/p}$
$\displaystyle\leq$
$\displaystyle\int_{y}^{\infty}\|f^{\prime}(\cdot+iv)\|_{p}dv$
$\displaystyle\lesssim$
$\displaystyle\int_{0}^{\infty}\frac{\|f^{\prime}(\cdot+iv)\|_{p}v^{1+\alpha}dv}{(y+v)^{1+\alpha}}$
$\displaystyle=$ $\displaystyle
H_{\alpha}\left(v\|f^{\prime}(\cdot+iv)\|_{p}\right)(y).$

The conclusion then follows from Theorem 2. ∎

###  4.3. Atomic decomposition of functions in
$A_{\omega^{k}}^{p,q}({\mathbb{C}}_{+})$

Let us start by recalling the following decomposition of ${\mathbb{C}}_{+}$
which is essentially from [8, Section 1].

####  4.3.1. A Whitney decomposition of ${\mathbb{C}}_{+}$

The Bergman distance in the upper-half plane is given by

$d_{Berg}(z,w):=\frac{1}{2}\log\left(\frac{1+\left|\frac{z-w}{\bar{z}-w}\right|}{1-\left|\frac{z-w}{\bar{z}-w}\right|}\right).$

We denote by $B_{\rho}(z_{0})$ the Bergman ball about $z_{0}$ with radius
$\rho$, i.e

$B_{\rho}(z_{0}):=\\{z\in{\mathbb{C}}_{+}:d_{Berg}(z,z_{0})<\rho\\}.$

Let $0<\delta<1$ and $\gamma$ a real such that

$\frac{\log\left(\frac{1+\frac{\delta^{2}}{20}}{1-\frac{\delta^{2}}{20}}\right)}{4\log
2}<\gamma<\frac{\log\left(\frac{1+\frac{\delta^{2}}{4}}{1-\frac{\delta^{2}}{4}}\right)}{4\log
2}.$

For $l,j\in\mathbb{Z}$, put

(4.4) $z_{l,j}=\frac{\delta^{2}}{4}l2^{\gamma j-1}+i2^{\gamma
j}=x_{l,j}+iy_{j},$

and define

(4.5) $\displaystyle I_{l,j}$
$\displaystyle:=\left\\{x\in{\mathbb{R}}:\left|2^{-\gamma
j}(x-x_{l,j})\right|<\frac{\delta^{2}}{4}\right\\}$ (4.6) $\displaystyle
I_{l,j}^{\prime}$ $\displaystyle:=\left\\{x\in{\mathbb{R}}:\left|2^{-\gamma
j}(x-x_{l,j})\right|<\frac{\delta^{2}}{20}\right\\}$ (4.7) $\displaystyle
J_{j}$ $\displaystyle:=\left\\{y\in(0,\infty):\left|2^{-\gamma
j}(y-y_{j})\right|<\frac{\delta^{2}}{4}\right\\}$ (4.8) $\displaystyle
J_{j}^{\prime}$ $\displaystyle:=\left\\{y\in(0,\infty):\left|2^{-\gamma
j}(y-y_{j})\right|<\frac{\delta^{2}}{20}\right\\}.$

We have also the following properties.

######  Lemma 5.

  * Let $l,j\in{\mathbb{Z}}$. The following assertions are satisfied.

  * (i)

$]0;\infty[=\cup_{j\in\mathbb{Z}}J_{j}$ and
$\mathbb{R}=\cup_{l\in\mathbb{Z}}I_{l,j}$, for all $j$ fixed in $\mathbb{Z}$;

  * (ii)

for fixed $j\in\mathbb{Z}$, intervals $I_{l,j}^{\prime}$ are pairwise
disjoint;

  * (iii)

the intervals $J_{j}^{\prime}$ are pairwise disjoint;

  * (iv)

for fixed $j\in\mathbb{Z}$, each element $x\in\mathbb{R}$ is in at most four
intervals $I_{l,j}$;

  * (v)

there exists an integer $N$ such that any $y\in]0;\infty[$ is in at most $N$
intervals $J_{j}$.

One can check (see [8]) that

$I_{l,j}+iJ_{j}\subset B_{\delta^{2}}(z_{l,j})$

and

$B_{\frac{\delta^{2}}{80}}(z_{l,j})\subset I_{l,j}^{\prime}+iJ_{j}^{\prime}.$

The sequence $\\{z_{l,j}\\}_{l,j\in{\mathbb{Z}}}$ above is called a
$\delta$-lattice in ${\mathbb{C}}_{+}$.

####  4.3.2. Atomic decomposition

For $y>0$ and $x\in\mathbb{R}$, put

(4.9) $L=L(y):=\\{j\in\mathbb{Z}:y\in J_{j}\\}$

and

(4.10) $L_{k}=L_{k}(x):=\\{l\in\mathbb{Z}:x\in I_{l,k}\\},$

for all $k\in\mathbb{Z}$.

In what follows, we will denote by $\mathcal{I}(F)$ the following quantity

$\int_{0}^{\infty}\sum_{j\in\mathbb{Z}}\chi_{J_{j}}(y)\left(\int_{0}^{\infty}\int_{-\infty}^{+\infty}|F(u+iv)|^{p}\left(\int_{-\infty}^{+\infty}\sum_{l\in\mathbb{Z}}\chi_{\\{x\in
I_{l,j}:|(x+iy)-(u+iv)|<\frac{y}{2\sqrt{2}}\\}}(x)dx\right)\frac{dudv}{v^{2}}\right)^{\frac{q}{p}}d\omega_{\alpha,k}(y),$

where $d\omega_{\alpha,k}(y)=\omega^{k}(y)y^{\alpha}dy$.

We have the following estimate.

######  Lemma 6.

Let $\alpha>-1$, $1\leq p,q<\infty$. Assume that
$\Phi\in\mathscr{L}\cup\mathscr{U}$. Put
$\omega=\mathcal{T}_{\Phi}^{\vec{\varepsilon}}$, and let $k\in\mathbb{R}$. For
$F\in A^{\Phi}_{\omega^{k},\alpha}(\mathbb{C_{+}})$,

(4.11) $\mathcal{I}(F)\leq C\|F\|_{A^{p,q}_{\omega^{k},\alpha}}^{q}.$

###### Proof.

From the property (iii) in Lemma 5, we have that

$\sum_{l\in\mathbb{Z}}\chi_{\\{x\in
I_{l,j}:|(x+iy)-(u+iv)|<\frac{y}{2\sqrt{2}}\\}}\leq 4.$

Hence

$\displaystyle S$ $\displaystyle:=$
$\displaystyle\int_{-\infty}^{+\infty}\sum_{l\in\mathbb{Z}}\chi_{\\{x\in
I_{l,j}:|(x+iy)-(u+iv)|<\frac{y}{2\sqrt{2}}\\}}(x)dx$ $\displaystyle\leq$
$\displaystyle
4\int_{\\{x\in{\mathbb{R}}:\,|(x+iy)-(u+iv)|<\frac{y}{2\sqrt{2}}\\}}dx$
$\displaystyle\leq$ $\displaystyle
4\chi_{\\{|y-v|<\frac{y}{2\sqrt{2}}\\}}(v)\int_{\\{x\in{\mathbb{R}}:\,|x-u|<\frac{y}{2\sqrt{2}}\\}}dx$
$\displaystyle\leq$ $\displaystyle
Cv\chi_{\\{|y-v|<\frac{y}{2\sqrt{2}}\\}}(v)$

where we have used in the last inequality the fact that if
$x+iy\in{\mathbb{C}}_{+}$ and $u+iv\in{\mathbb{C}}_{+}$ are such that
$|(x+iy)-(u+iv)|<\frac{y}{2\sqrt{2}}$, then $y\simeq v$. It follows that,

$\displaystyle\mathcal{I}(F)$
$\displaystyle\lesssim\int_{0}^{\infty}\sum_{j\in\mathbb{Z}}\chi_{J_{j}}(y)\left(\int_{0}^{\infty}\int_{-\infty}^{+\infty}|F(u+iv)|^{p}\left(v\chi_{\\{y/C<v<Cy\\}}(v)\right)\frac{dudv}{v^{2}}\right)^{\frac{q}{p}}d\omega_{\alpha,k}(y)$
$\displaystyle\lesssim\int_{0}^{\infty}\sum_{j\in\mathbb{Z}}\chi_{J_{j}}(y)\left(\int_{y/C}^{Cy}\int_{-\infty}^{+\infty}|F(u+iv)|^{p}v\frac{dudv}{v^{2}}\right)^{\frac{q}{p}}d\omega_{\alpha,k}(y)$
$\displaystyle=\int_{0}^{\infty}\sum_{j\in\mathbb{Z}}\chi_{J_{j}}(y)\left(\int_{y/C}^{Cy}\|F(.+iv)\|_{L^{p}}^{p}\frac{dv}{v}\right)^{\frac{q}{p}}d\omega_{\alpha,k}(y).$

Given that $y\mapsto\|F(.+iy)\|_{L^{p}}$ is non-increasing on $(0,\infty)$, we
have that

$\int_{y/C}^{Cy}\|F(.+iv)\|_{L^{p}}^{p}\frac{dv}{v}\lesssim\left\|F\left(.+i\frac{y}{C}\right)\right\|_{L^{p}}^{p}.$

Therefore, using property (v) of Lemma 5, we obtain

$\displaystyle\mathcal{I}(F)$
$\displaystyle\lesssim\int_{0}^{\infty}\sum_{j\in\mathbb{Z}}\chi_{J_{j}}(y)\left(\int_{y/C}^{Cy}\|F(.+iv)\|_{L^{p}}^{p}\frac{dv}{v}\right)^{\frac{q}{p}}d\omega_{\alpha,k}(y)$
$\displaystyle\lesssim\int_{0}^{\infty}\sum_{j\in
L}\chi_{J_{j}}(y)\left\|F\left(.+i\frac{y}{C}\right)\right\|_{L^{p}}^{q}d\omega_{\alpha,k}(y)$
$\displaystyle\lesssim\int_{0}^{\infty}\left\|F\left(.+iy\right)\right\|_{L^{p}}^{q}d\omega_{\alpha,k}(y)\lesssim\|F\|_{A^{p,q}_{\omega^{k},\alpha}}^{q}.$

∎

The following result follows essentially from the Mean Value Theorem, we refer
to [8] for details.

######  Lemma 7.

For $1\leq p<\infty$, there exists a constant $C>0$ such that for every
holomorphic function $F$ on $\mathbb{C}_{+}$, and for every $\delta\in(0,1)$,
the following hold.

  * (a)

$|F(z)|^{p}\leq
C\delta^{-2}\int_{d_{Berg}(z,u+iv)<\delta}|F(u+iv)|^{p}\frac{dudv}{v^{2}}$.

  * (b)

If $d_{Berg}(z,\zeta)<\delta$ and $\delta$ is small enough, then

$|F(z)-F(\zeta)|^{p}\leq
C\delta^{p}\int_{d_{Berg}(z,u+iv)<1}|F(u+iv)|^{p}\frac{dudv}{v^{2}}.$

Let us prove the following lemma.

######  Lemma 8.

Suppose that $\delta\in(0,1)$, and let $1\leq p,q<\infty$. Then there exists a
constant $C>0$ such that the following hold.

(4.12) $\|F(\cdot+iy)\|_{p}^{q}\leq
C\int_{|v-y|<y\frac{\delta^{2}}{4}}\|F(\cdot+iv)\|_{p}^{q}\frac{dv}{v}.$

###### Proof.

The Mean Value Theorem gives us that

$|F(x+i)|^{p}\leq
C_{p}\int_{|v-1|<\frac{\delta^{2}}{4}}\int_{|x-u|<\frac{\delta^{2}}{4}}|F(u+iv)|^{p}dudv.$

It follows that

$\displaystyle\int_{\mathbb{R}}|F(x+i)|^{p}dx$ $\displaystyle\leq$
$\displaystyle
C_{p}\int_{|v-1|<\frac{\delta^{2}}{4}}\int_{|s|<\frac{\delta^{2}}{4}}\int_{\mathbb{R}}|F(x+iv)|^{p}dxdsdv$
$\displaystyle\leq$ $\displaystyle
C_{p}\frac{\delta^{2}}{4}\int_{|v-1|<\frac{\delta^{2}}{4}}\int_{\mathbb{R}}|F(x+iv)|^{p}dxdv.$

Using Jensen’s inequality when $q<p$ or H’́older’s inequality when $p\leq q$,
we obtain

(4.13) $\|F(\cdot+i)\|_{p}^{q}\leq
C_{p,q,\delta}\int_{|v-1|<\frac{\delta^{2}}{4}}\left(\int_{\mathbb{R}}|F(x+iv)|^{p}dx\right)^{\frac{q}{p}}dv.$

Let us apply (4.13) to the function $F_{y}$ defined by $F_{y}(u+iv)=F(u+iyv)$.
Using that for $v\in\\{v>0:|v-y|<y\frac{\delta^{2}}{4}\\}$, $v\simeq y$, we
obtain

$\displaystyle\|F(\cdot+iy)\|_{p}^{q}$ $\displaystyle=$
$\displaystyle\|F_{y}(\cdot+i)\|_{p}^{q}$ $\displaystyle\lesssim$
$\displaystyle\int_{|v-1|<\frac{\delta^{2}}{4}}\left(\int_{\mathbb{R}}|F_{y}(x+iv)|^{p}dx\right)^{\frac{q}{p}}dv$
$\displaystyle=$
$\displaystyle\int_{|v-1|<\frac{\delta^{2}}{4}}\left(\int_{\mathbb{R}}|F(x+iyv)|^{p}dx\right)^{\frac{q}{p}}dv$
$\displaystyle\lesssim$
$\displaystyle\int_{|v-y|<y\frac{\delta^{2}}{4}}\|F(\cdot+iv)\|_{p}^{q}\frac{dv}{v}.$

The proof is complete. ∎

######  Theorem 6.

Let $\alpha>-1$, $0<\delta<1$ and $\\{z_{l,j}=\frac{\delta^{2}}{4}l2^{\gamma
j-1}+i2^{\gamma j}\\}_{l,j\in{\mathbb{Z}}}$ the associated $\delta$-lattice
defined above. Let $\alpha>-1$ and $1<p,q<\infty$. Assume that
$\Phi\in\mathscr{L}\cup\mathscr{U}$. Put
$\omega=\mathcal{T}_{\Phi}^{\vec{\varepsilon}}$, and let $k\in\mathbb{R}$.
There exists a $C_{\delta}=C(\delta,p,q)>0$ such that for every $F\in
A^{p,q}_{\omega^{k}\alpha}(\mathbb{C_{+}})$,

(4.14)
$\sum_{j}\left(\sum_{l}|F(z_{l,j})|^{p}\right)^{\frac{q}{p}}\omega^{k}(2^{\gamma
j})2^{\gamma j(\alpha+1+\frac{q}{p})}\leq
C_{\delta}\|F\|_{A^{p,q}_{\omega^{k},\alpha}}^{q}.$

Moreover, if $\delta$ is small enough, then the converse inequality holds, i.e

(4.15) $\|F\|_{A^{p,q}_{\omega^{k},\alpha}}^{q}\leq
C_{\delta}\sum_{j}\left(\sum_{l}|F(z_{l,j})|^{p}\right)^{\frac{q}{p}}\omega^{k}(2^{\gamma
j})2^{\gamma j(\alpha+1+\frac{q}{p})}.$

###### Proof.

From assertion (a) in Lemma 7 and the fact that
$B_{\frac{\delta^{2}}{80}}(z_{l,j})\subset I_{l,j}^{\prime}+iJ_{j}^{\prime}$,
we obtain that

$\displaystyle|F(z_{l,j})|^{p}$ $\displaystyle\lesssim$
$\displaystyle\delta^{-4}\int_{B_{\frac{\delta^{2}}{80}}(z_{l,j})}|F(u+iv)|^{p}\frac{dudv}{v^{2}}$
$\displaystyle\leq$
$\displaystyle\delta^{-4}\int_{I_{l,j}^{\prime}}du\int_{J_{j}^{\prime}}|F(u+iv)|^{p}\frac{dv}{v^{2}}$
$\displaystyle\lesssim$
$\displaystyle\frac{\delta^{-4}}{y_{j}^{2}}\int_{I_{l,j}^{\prime}}du\int_{J_{j}^{\prime}}|F(u+iv)|^{p}dv$

where we have used the fact that for $v\in J_{j}^{\prime}$, $v\simeq y_{j}$.
It follows that

$\displaystyle\sum_{l\in\mathbb{Z}}|F(z_{l,j})|^{p}$ $\displaystyle\lesssim$
$\displaystyle\sum_{l\in\mathbb{Z}}\frac{\delta^{-4}}{y_{j}^{2}}\int_{I_{l,j}^{\prime}}du\int_{J_{j}^{\prime}}|F(u+iv)|^{p}dv$
$\displaystyle\lesssim$
$\displaystyle\frac{\delta^{-4}}{y_{j}^{2}}\int_{\mathbb{R}}du\int_{J_{j}^{\prime}}|F(u+iv)|^{p}dv$
$\displaystyle=$
$\displaystyle\frac{\delta^{-4}}{y_{j}^{2}}\int_{J_{j}^{\prime}}\|F(\cdot+iv)\|_{p}^{p}dv$
$\displaystyle\leq$
$\displaystyle\frac{\delta^{-4}}{y_{j}^{2}}\int_{J_{j}^{\prime}}\|F(\cdot+i\lambda
y_{j})\|_{p}^{p}dv\quad\left(\lambda y_{j}<v\right)$ $\displaystyle\lesssim$
$\displaystyle\frac{\delta^{-4}}{y_{j}}\|F(\cdot+i\lambda y_{j})\|_{p}^{p}.$

Hence using Lemma 8, we obtain

$\displaystyle\sum_{j\in\mathbb{Z}}\left(\sum_{l\in\mathbb{Z}}|F(z_{l,j})|^{p}\right)^{\frac{q}{p}}\omega^{k}(2^{\gamma
j})2^{\gamma j(\alpha+1+\frac{q}{p})}$ $\displaystyle\leq$ $\displaystyle
C_{\delta}\sum_{j\in\mathbb{Z}}\|F(\cdot+i\lambda
y_{j})\|_{p}^{q}\omega^{k}(y_{j})y_{j}^{\alpha+1}$ $\displaystyle\lesssim$
$\displaystyle
C_{\delta}\sum_{j\in\mathbb{Z}}\int_{J_{j}^{\prime}}\|F(\cdot+iy)\|_{p}^{q}\omega^{k}(y_{j})y^{\alpha}dy$
$\displaystyle\lesssim$ $\displaystyle
C_{\delta}\int_{0}^{\infty}\|F(\cdot+iy)\|_{p}^{q}\omega^{k}(y)y^{\alpha}dy.$

To prove the converse, we first note that

$\displaystyle\|F(\cdot+iy)\|_{p}^{p}$ $\displaystyle\leq$ $\displaystyle
C_{p}\left\\{\sum_{l\in\mathbb{Z}}\int_{I_{l,j}}|F(x+iy)-F(z_{l,j})|^{p}dx+\sum_{l\in\mathbb{Z}}\int_{I_{l,j}}|F(z_{l,j})|^{p}dx\right\\}$
$\displaystyle\leq$ $\displaystyle
C_{p,\delta}\left\\{\sum_{l\in\mathbb{Z}}\int_{I_{l,j}}|F(x+iy)-F(z_{l,j})|^{p}dx+\sum_{l\in\mathbb{Z}}|F(z_{l,j})|^{p}y_{j}\right\\}.$

Using assertion (b) in Lemma 7, we obtain

$\displaystyle T(y)$ $\displaystyle:=$
$\displaystyle\sum_{l\in\mathbb{Z}}\int_{I_{l,j}}|F(x+iy)-F(z_{l,j})|^{p}dx$
$\displaystyle=$
$\displaystyle\sum_{l\in\mathbb{Z}}\int_{\mathbb{R}}|F(x+iy)-F(z_{l,j})|^{p}\chi_{I_{l,j}}(x)dx$
$\displaystyle\leq$ $\displaystyle
C\delta^{p}\int_{0}^{\infty}\int_{-\infty}^{\infty}|F(u+iv)|^{p}\left(\int_{\mathbb{R}}\sum_{l\in
L_{j}}\chi_{x\in I_{l,j}:|u+iv-x-
iy|<C_{\delta}y}(x)dx\right)\frac{dudv}{v^{2}}.$

It follows that

$\displaystyle\|F\|_{A^{p,q}_{\omega^{k},\alpha}}^{q}$ $\displaystyle\leq$
$\displaystyle
Cq\int_{0}^{\infty}\sum_{j\in\mathbb{Z}}\chi_{J_{j}}(y)T(y)^{\frac{q}{p}}\omega^{k}(y)y^{\alpha}dy+\sum_{j\in\mathbb{Z}}\left(\sum_{l\in\mathbb{Z}}|F(z_{l,j})|^{p}\right)^{\frac{q}{p}}\omega^{k}(y_{j})y_{j}^{(\alpha+1+\frac{q}{p})}$
$\displaystyle\lesssim$
$\displaystyle\delta^{q}\mathcal{I}(F)+\sum_{j\in\mathbb{Z}}\left(\sum_{l\in\mathbb{Z}}|F(z_{l,j})|^{p}\right)^{\frac{q}{p}}\omega^{k}(y_{j})y_{j}^{(\alpha+1+\frac{q}{p})}.$

Using Lemma 6, we obtain

$\|F\|_{A^{p,q}_{\omega^{k},\alpha}}^{q}\leq
C\left(\delta^{q}\|F\|_{A^{p,q}_{\omega^{k},\alpha}}^{q}+\sum_{j\in\mathbb{Z}}\left(\sum_{l\in\mathbb{Z}}|F(z_{l,j})|^{p}\right)^{\frac{q}{p}}\omega^{k}(y_{j})y_{j}^{(\alpha+1+\frac{q}{p})}\right).$

This yields the inequality (4.15) for $\delta$ small enough. ∎

For $\alpha>-1$, $1\leq p,q<\infty$ and $k\in\mathbb{R}$, we define the
complex sequences space
$\ell_{\omega^{k},\alpha}^{p,q}=\ell_{\omega^{k},\alpha}^{p,q}({\mathbb{C}}_{+})$
by

$\ell_{\omega^{k},\alpha}^{p,q}:=\left\\{\lambda=\\{\lambda_{l,j}\\}_{l,j\in{\mathbb{Z}}}:\|\lambda\|_{\ell_{\omega^{k},\alpha}^{p,q}}=\left(\sum_{j\in{\mathbb{Z}}}\left(\sum_{l\in{\mathbb{Z}}}|\lambda_{l,j}|^{p}\right)^{\frac{q}{p}}\omega^{k}(2^{\gamma
j})2^{\gamma j(\alpha+1+\frac{q}{p})}\right)^{\frac{1}{q}}<\infty\right\\}.$

We refer to [8] for the following.

######  Lemma 9.

Let $\alpha>-1$ and $1<p,q<\infty$. Assume that
$\Phi\in\mathscr{L}\cup\mathscr{U}$. Put
$\omega=\mathcal{T}_{\Phi}^{\vec{\varepsilon}}$, and let $k\in\mathbb{R}$. The
topological dual of $\ell_{\omega^{k},\alpha}^{p,q}({\mathbb{C}}_{+})$,
$\left(\ell_{\omega^{k},\alpha}^{p,q}({\mathbb{C}}_{+})\right)^{*}$ coincides
with
$\ell_{\omega^{k(1-q^{\prime})},\alpha}^{p^{\prime},q^{\prime}}({\mathbb{C}}_{+})$,
in the sense that, for
$T\in\left(\ell_{\omega^{k},\alpha}^{p,q}({\mathbb{C}}_{+})\right)^{*}$, there
is a unique
$\mu:=\\{\mu_{l,j}\\}_{l,j\in\mathbb{Z}}\in\ell_{\omega^{k(1-q^{\prime})},\alpha}^{p^{\prime},q^{\prime}}({\mathbb{C}}_{+})$
such that for all
$\lambda:=\\{\lambda_{l,j}\\}_{l,j\in\mathbb{Z}}\in\ell_{\omega^{k},\alpha}^{p,q}({\mathbb{C}}_{+})$,

(4.16)
$T(\mu,\lambda):=\langle\mu,\lambda\rangle=\sum_{j\in\mathbb{Z}}\sum_{l\in\mathbb{Z}}\mu_{j,l}\overline{\lambda_{j,l}}2^{j\gamma(\alpha+2)}.$

We have the following atomic decomposition of the weighted Bergman spaces in
consideration.

######  Theorem 7.

Let $\alpha>-1$, $1<p,q<\infty$, $0<\delta<1$ and
$\\{z_{l,j}=\frac{\delta^{2}}{4}l2^{\gamma j-1}+i2^{\gamma
j}\\}_{l,j\in{\mathbb{Z}}}$ the associated $\delta$-lattice. Assume that
$\Phi\in\mathscr{L}\cup\mathscr{U}$. Put
$\omega=\mathcal{T}_{\Phi}^{\vec{\varepsilon}}$, and let $k\in\mathbb{R}$.
Then the following assertions are satisfied.

  * (a)

For any sequence of complex numbers $\\{\lambda_{l,j}\\}_{l,j\in{\mathbb{Z}}}$
such that

$\sum_{j\in{\mathbb{Z}}}\left(\sum_{l\in{\mathbb{Z}}}|\lambda_{l,j}|^{p}\right)^{\frac{q}{p}}\omega^{k}(2^{\gamma
j})2^{\gamma j(\alpha+1+\frac{q}{p})}<\infty,$

the series

$\sum_{l,j}\lambda_{l,j}2^{\gamma
j(\alpha+1+\frac{q}{p})}K_{\alpha}(z,z_{l,j})$

is convergent in $A^{p,q}_{\omega^{k},\alpha}({\mathbb{C}}_{+})$. Moreover,
its sum $F$ satisfies the estimate

(4.17) $\|F\|_{A^{p,q}_{\omega^{k},\alpha}}^{q}\leq
C_{\delta}\sum_{j}\left(\sum_{l}|F(z_{l,j})|^{p}\right)^{\frac{q}{p}}\omega^{k}(2^{\gamma
j})2^{\gamma j(\alpha+1+\frac{q}{p})}.$

  * (b)

If $\delta$ is small enough, then any function $F\in
A^{p,q}_{\omega^{k},\alpha}({\mathbb{C}}_{+})$ may be represented as

$F(z)=\sum_{l,j}\lambda_{l,j}2^{\gamma
j(\alpha+1+\frac{q}{p})}K_{\alpha}(z,z_{l,j})$

with

(4.18)
$\sum_{j}\left(\sum_{l}|F(z_{l,j})|^{p}\right)^{\frac{q}{p}}\omega^{k}(2^{\gamma
j})2^{\gamma j(\alpha+1+\frac{q}{p})}\leq
C_{\delta}\|F\|_{A^{p,q}_{\omega^{k},\alpha}}^{q}.$

###### Proof.

The follows the now classical arguments in the literature (see for example
[5]).

We recall with Lemma 7 that the dual space
$\left(A_{\omega^{k},\alpha}^{p,q}(\mathbb{C}_{+})\right)^{*}$ of the Bergman
space $A_{\omega^{k},\alpha}^{p,q}(\mathbb{C}_{+})$ identifies with
$A_{\omega^{k(1-q^{\prime})},\alpha}^{p^{\prime},q^{\prime}}(\mathbb{C}_{+})$
under the integral pairing

$\langle
f,g\rangle_{\alpha}=\int_{\mathbb{C}_{+}}f(z)\overline{g(z)}dV_{\alpha}(z).$

Lemma 9 gives us that the dual space
$(\ell_{\omega^{k},\alpha}^{p,q}(\mathbb{C}))^{*}$ of
$\ell_{\omega^{k},\alpha}^{p,q}(\mathbb{C})$ identifies with
$\ell_{\omega^{k(1-q^{\prime})},\alpha}^{p^{\prime},q^{\prime}}(\mathbb{C})$
under the sum pairing (4.16).

$(i)$ From the first part of the sampling theorem, Theorem 6, we deduce that
the linear operator

$R:\,\,\,A_{\omega^{k},\alpha}^{p,q}(\mathbb{C}_{+})\rightarrow\ell_{\omega^{k},\alpha}^{p,q}(\mathbb{C})$
${F\mapsto\\{F(z_{l,j})\\}}$

is bounded. Hence its adjoint under the pairing
$\langle\cdot,\cdot\rangle_{\alpha}$

$R^{*}:\,\,\,\ell_{\omega^{k(1-q^{\prime})},\alpha}^{p^{\prime},q^{\prime}}(\mathbb{C})\rightarrow
A_{\omega^{k(1-q^{\prime})},\alpha}^{p^{\prime},q^{\prime}}(\mathbb{C}_{+})$

is also bounded. It is then enough to show that

$R^{*}(\\{\lambda_{l,j}\\})=\sum_{l,j}{\lambda_{l,j}2^{\gamma
j(\alpha+2)}K_{\alpha}(\cdot,z_{l,j})}.$

For $F\in A_{\omega^{k},\alpha}^{p,q}(\mathbb{C}_{+})$ and
$\\{\lambda_{l,j}\\}\in\ell_{\omega^{k},\alpha}^{p,q}(\mathbb{C})$, we have

$\displaystyle\langle RF,\\{\lambda_{l,j}\\}\rangle_{\alpha}$ $\displaystyle=$
$\displaystyle\sum_{l,j}F(z_{l,j})\overline{\lambda_{l,j}}2^{j\gamma(\alpha+2)}$
$\displaystyle=$
$\displaystyle\sum_{l,j}(P_{\alpha}F(z_{l,j}))\overline{\lambda_{l,j}}2^{j\gamma(\alpha+2)}$
$\displaystyle=$
$\displaystyle\sum_{l,j}\int_{\mathbb{C}_{+}}K_{\alpha}(z_{l,j},w)F(w)dV_{\alpha}(w)\overline{\lambda_{l,j}}2^{j\gamma(\alpha+2)}$
$\displaystyle=$
$\displaystyle\int_{\mathbb{C}_{+}}F(w)\overline{\left(\sum_{l,j}{\lambda_{l,j}}K_{\alpha}(w,z_{l,j})2^{j\gamma(\alpha+2)}\right)}dV_{\alpha}(w)$
$\displaystyle=$ $\displaystyle\langle
F,R^{*}(\\{\lambda_{l,j}\\})\rangle_{\alpha}.$

This conclude the first part of the theorem.

$(ii)$ From the second part of Theorem 6, we have for $\delta$ small enough,

$||F||_{A_{\omega^{k(1-q^{\prime})},\alpha}^{p^{\prime},q^{\prime}}}\leq
C_{\delta}||\\{F(z_{l,j})\\}||_{\ell_{\omega^{k(1-q^{\prime})},\alpha}^{p^{\prime},q^{\prime}}}.$

Thus
$R^{*}:\,\,\,\ell_{\omega^{k(1-q^{\prime})},\alpha}^{p^{\prime},q^{\prime}}(\mathbb{C})\rightarrow
A_{\omega^{k(1-q^{\prime})},\alpha}^{p^{\prime},q^{\prime}}(\mathbb{C}_{+})$
is onto. Moreover, if we denote by $\mathcal{N}$ the subspace of
$\ell_{\omega^{k(1-q^{\prime})},\alpha}^{p^{\prime},q^{\prime}}(\mathbb{C})$
consisting of all sequences $\\{\lambda_{l,j}\\}_{l,j\in\mathbb{Z}}$ such that
the sum $\sum_{l,j}\lambda_{l,j}K_{\alpha}(z,z_{l,j})2^{\gamma
j(\alpha+1+\frac{q}{p})}$ is identically zero, then the linear map

$\ell_{\omega^{k(1-q^{\prime})},\alpha}^{p^{\prime},q^{\prime}}(\mathbb{C})\setminus\mathcal{N}\rightarrow
A_{\omega^{k(1-q^{\prime})},\alpha}^{p^{\prime},q^{\prime}}(\mathbb{C}_{+})$

is a bounded isomorphism. The continuity of its inverse which follows from the
Hahn-Banach theorem gives the estimate (4.18). This completes the proof of the
theorem.

∎

## References

  * [1] Bansah, J., Sehba, B.: Boundedness of a family of Hilbert-type operators and its Bergman-type analogue. Illinois J. Math. 59 (2015), no. 4, 949–977.
  * [2] Bèkollè, D.: Inégalités à poids pour le projecteur de Bergman dans la boule unité de ${\mathbb{C}}^{n}$. Stud. Math. (1982) Vol 71, 305–323.
  * [3] Bekollé, D., Bonami, A.: Inégalités à poids pour le noyau de Bergman. (French) C. R. Acad. Sci. Paris Sér. A-B 286 (1978), no. 18, A775–A778.
  * [4] Bèkollè, D., Bonami, A., Garrigós, G., Peloso, M., Ricci, F.: Boundedness of Bergman projections on tube domains over light cones. Math. Z. 237 (2001), no. 1, 31-59.
  * [5] Bèkollè, D., Bonami, A., Garrigós, G., Nana, C., Peloso, M., Ricci, F.: Lecture notes on Bergman projectors in tube domains over cones: an analytic and geometric viewpoint. IMHOTEP J. Afr. Math. Pures Appl. 5 (2004), Exp. I, front matter + ii + 75 pp
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  * [7] García-Cuerva, J., Rubio de Francia, J. L., Weighted norm inequalities and related topics. North-Holland Mathematics Studies, 116. Notas de Matemática [Mathematical Notes], 104. North-Holland Publishing Co., Amsterdam, 1985. x+604 pp. ISBN: 0-444-87804-1
  * [8] Gonessa, J., Espaces de type Bergman dans les domaines homogènes de Siegel de type II: Décomposition atomique et interpolation. PhD 2006. University of Yaoundé I.
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  * [10] Sehba, B. F.: Bergman type operators in tubular domains over symmetric cones, Proc. Edin. Math. Soc. 52 (2) (2009), 529-544. $H^{1}$ and the functions of bounded mean oscillation, Amer. J. Math.
  * [11] Stein, E. M. & Weiss, G.: Introduction to Fourier Analysis on Euclidean Spaces. Princeton Math. Ser., No. 32 Princeton University Press, Princeton, NJ, 1971, x+297 pp


# Improving Breast Cancer Grade Prediction with Multiparametric MRI Created
Using Optimized Synthetic Correlated Diffusion Imaging

Chi-en Amy Tai
University of Waterloo
Waterloo, ON
<EMAIL_ADDRESS>Alexander Wong
University of Waterloo
Waterloo, ON
<EMAIL_ADDRESS>
###### Abstract

Breast cancer was diagnosed for over 7.8 million women between 2015 to 2020.
Grading plays a vital role in breast cancer treatment planning. However, the
current tumor grading method involves extracting tissue from patients, leading
to stress, discomfort, and high medical costs. A recent paper leveraging
volumetric deep radiomic features from synthetic correlated diffusion imaging
(CDIs) for breast cancer grade prediction showed immense promise for
noninvasive methods for grading. Motivated by the impact of CDIs optimization
for prostate cancer delineation, this paper examines using optimized CDIs to
improve breast cancer grade prediction. We fuse the optimized CDIs signal with
diffusion-weighted imaging (DWI) to create a multiparametric MRI for each
patient. Using a larger patient cohort and training across all the layers of a
pretrained MONAI model, we achieve a leave-one-out cross-validation accuracy
of 95.79%, over 8% higher compared to that previously reported.

##  1 Introduction

Breast cancer was diagnosed for over 7.8 million women between 2015 to 2020
and resulted in 685,000 deaths across the world in 2020 [10]. Grading is a
crucial factor in breast cancer treatment planning, but the current method to
grade breast cancer tumors involves tissue extraction from the patient which
causes patient stress and discomfort [14], along with high medical fees [7].
In addition, the current method to determine the cancer grade is based on a
human pathologist’s opinion [1]. With potential biases and high uncertainty in
clinical judgement, it is possible for the patient to receive an incorrect
grade leading to an unsuitable treatment strategy [12]. Recent advancements in
computer vision and imaging in the cancer domain have shown promise of
noninvasive ways to diagnose and evaluate cancer tumours with high accuracy
[13].

In literature, multiparametric MRIs have recently gained popularity as they
have been shown to greatly benefit breast cancer clinical task enhancement
with deep learning [6]. Multiparametric MRIs are created by fusing two or more
MRI modalities together with the combination achieving better performance than
each MRI modality on its own [5].

Specifically, leveraging volumetric deep radiomic features from synthetic
correlated diffusion imaging (CDIs) fused with DWI for breast cancer grade
prediction showed promise compared to other MRI modalities, achieving a
prediction accuracy of 87.70% [13]. First introduced in Wong et al. [15], CDIs
is a method that utilizes a combination of native and synthetic diffusion
signal acquisitions, coupled with a signal calibration that aims to achieve
enhanced consistency in the dynamic range across various machines and
protocols. CDIs, on its own, was shown to produce superior results for
prostate cancer delineation compared to other gold-standard modalities.

In Wong et al. [15], they demonstrate the impact of tuning CDIs for the
specific cancer domain. However, in Tai et al. [13], they only slightly
modified CDIs for breast cancer instead of properly tuning the coefficients.
Motivated by the impact of CDIs optimization for prostate cancer delineation
in [15], this paper examines using optimized CDIs to improve breast cancer
grade prediction.

##  2 Methodology

Table 1: SBR grade distribution in the patient cohort.

l c SBR Grade Number of Patients
Grade I (Low) 10
Grade II (Intermediate) 99
Grade III (High) 200

Figure 1: Clinical support workflow copied from [13] that is used in this
study.

In this study, we use the American College of Radiology Imaging Network
(ACRIN) 6698/I-SPY2 study [11, 8, 9, 3] and filter for non-null Scarff-Bloom-
Richardson (SBR) grade values with a total of 309 patients remaining. Similar
to [13], we also combine SBR grade I and II into one category due to the
imbalanced distribution between the three grades (as shown in Table 2).

To obtain the optimized CDIs coefficients, we leverage the Nelder-Mead simplex
optimization strategy to maximize the area under the receiver operating
characteristic curve (AUC) for breast cancer tumour delineation. After
generating the optimized CDIs signals, we fuse the signals on the diffusion-
weighted imaging (DWI) to create a multiparametric MRI. To achieve dimensional
consistency for machine learning, all volumes are then standardized into
224x224x25 volumetric data cubes for each patient.

Using a previously introduced deep radiomic clinical support workflow [13]
(shown in Figure 1), we leverage a pretrained 34-layer volumetric residual
convolutional neural network architecture initialized with weights from MONAI
[4] and train it to extract deep radiomic features. The MONAI weights were
derived by training on the extensive 3D medical dataset, 3DSeg-8. This
comprehensive dataset comprises images from eight different 3D segmentation
datasets, encompassing both MRI and CT images [2]. These features are then fed
into a fully-connected neural network grade predictor to predict breast cancer
grade (Grade I/Grade II and Grade III).

For training, a learning rate of 0.001 was used along with a weighted random
sampler, AdamW optimizer, cross-entropy loss function and cosine annealing
learning rate scheduler. We also train all the model layers with no freezing.
Leave-one-out cross-validation was conducted to obtain the results with the
average accuracy, sensitivity, and specificity recorded.

##  3 Results

As seen in Table 3, using the optimized CDIs obtained a leave-one-out cross-
validation of 95.79%, over 8% higher than that previously reported (87.70%).
Optimizing CDIs also achieved higher sensitivity and specificity, with all
values above 90%. Figure 2 shows an illustrative example highlighting the
visual differences between an unoptimized CDIs signal and an optimized CDIs
signal, along with the corresponding tumour mask and DWI. Given the promising
results, this proposed noninvasive method to identify the severity of cancer
would allow for better treatment decisions without the need for a biopsy and
highlights the impact of optimizing CDIs for the specific cancer domain.

Table 2: Results using unoptimized and optimized CDIs with the best results
bolded.

l c c c CDIs Version Accuracy Sensitivity Specificity
Unoptimized [13] 87.70% 90.29% 81.82%
Optimized 95.79% 96.50% 94.50%

(a)

(b)

(c)

(d)

Figure 2: An example slice illustrating visual differences between (a)
Unoptimized CDIs, (b) Optimized CDIs, (c) the associated DWI, and (d) the
associated tumour mask for a patient who has SBR Grade III (High). In this
patient case, grade prediction was correct using the Optimized CDIs signal
fused with DWI.

## References

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