$\displaystyle\qquad w_{2}^{A_{\gamma}}(\zeta)$
$\displaystyle=\frac{\zeta_{2}-i\gamma|\zeta_{1}|^{\gamma}}{1-\gamma}$
$\displaystyle=\frac{\left(1-\frac{\gamma}{2}\right)\zeta_{2}+\frac{\gamma}{2}\overline{\zeta}_{2}}{1-\gamma}.$

This shows that the defining equation for the dual hypersurface
$M_{\gamma}^{*,A_{\gamma}}$ is

(6.12)
$\text{Im}\left(w_{2}^{A_{\gamma}}(\zeta)\right)=\left|w_{1}^{A_{\gamma}}(\zeta)\right|^{\tfrac{\gamma}{\gamma-1}}.$

Thus via the matrix $A_{\gamma}$, the dual hypersurface
$M_{\gamma}^{*}=M_{\gamma^{*}}$, where $\gamma^{*}=\tfrac{\gamma}{\gamma-1}$
is the Hölder conjugate of the exponent $\gamma$.

Note for future reference that

(6.13a) $\displaystyle\zeta_{1}$
$\displaystyle=\overline{w_{1}^{A_{\gamma}}}\left|w_{1}^{A_{\gamma}}\right|^{\frac{2-\gamma}{\gamma-1}}$
(6.13b) $\displaystyle|\zeta_{1}|$
$\displaystyle=\left|w_{1}^{A_{\gamma}}\right|^{\frac{1}{\gamma-1}}$ (6.13c)
$\displaystyle\zeta_{2}$
$\displaystyle=\left(1-\frac{\gamma}{2}\right)w_{2}^{A_{\gamma}}-\frac{\gamma}{2}\overline{w_{2}^{A_{\gamma}}}=(1-\gamma)w_{2}^{A_{\gamma}}+i\gamma\left|w_{1}^{A_{\gamma}}\right|^{\frac{\gamma}{\gamma-1}}$
(6.13d) $\displaystyle\operatorname{Re}\zeta_{2}$
$\displaystyle=(1-\gamma)\operatorname{Re}w_{2}^{A_{\gamma}}.$

######  Remark 6.14.

Using the $(s,\alpha,\theta)$ parametrization of $M_{\gamma}$ from Section
3.2, the matrix in (6.2c) has determinant

$\Delta:=P+Qe^{-i\theta}\alpha^{\gamma-1}+R\left(\alpha^{\gamma}+i\frac{s}{\gamma-1}\right)$

where

$\displaystyle P$
$\displaystyle=\frac{i}{2}\left(b_{1}m_{21}-b_{2}m_{11}\right)$ $\displaystyle
Q$ $\displaystyle=\frac{\gamma}{2}\left(b_{2}m_{12}-b_{1}m_{22}\right)$
$\displaystyle R$
$\displaystyle=\frac{\gamma-1}{2}\left(m_{12}m_{21}-m_{11}m_{22}\right).$

Thus, the matrix $A$ is admissible if and only if $\Delta$ is non-zero for all
$s,\theta\in{\mathbb{R}},\alpha>0$. We make the following observations.

  * •

Invertibility of $A$ implies that $P,Q,R$ are not all zero.

  * •

If $R=0$ then it is easy to see that $M_{\gamma}$ is $A$-admissible if and
only if one of $P,Q$ vanishes.

  * •

For $R\neq 0$ we can apply Young’s inequality in the form
$\alpha^{\gamma-1}\leq\frac{1}{\gamma}+\frac{\gamma-1}{\gamma}\alpha^{\gamma}$
to obtain

$\displaystyle\operatorname{Re}\Delta\overline{R}$
$\displaystyle=\operatorname{Re}P\overline{R}+R\overline{R}\alpha^{\gamma}+\operatorname{Re}(Q\overline{R}e^{-i\theta})\cdot\alpha^{\gamma-1}$
$\displaystyle\geq\left(\operatorname{Re}P\overline{R}-\frac{|QR|}{\gamma}\right)+\left(|R|^{2}-\frac{(\gamma-1)|QR|}{\gamma}\right)\alpha^{\gamma};$

so if
$|Q|\leq\gamma\min\left\\{\operatorname{Re}\frac{P\overline{R}}{|R|},\frac{|R|}{\gamma-1}\right\\}$
(with strict inequality if
$\operatorname{Re}\frac{P\overline{R}}{|R|}=\frac{|R|}{\gamma-1}$) then
$\operatorname{Re}\Delta\overline{R}$ is strictly positive and thus
$M_{\gamma}$ is $A$-admissible.

  * •

On the other hand, if $Q=0$ and $P\overline{R}$ is negative then it is easy to
see that $M_{\gamma}$ is not $A$-admissible. $\lozenge$

######  Remark 6.15.

In general, when a hypersurface ${\mathcal{S}}$ is both $A$-admissible and
$A^{\prime}$-admissible then the dualization maps $w^{A}$ and $w^{A^{\prime}}$
are related by
$w^{A^{\prime}}=w^{A}\circ\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[width("\left({A(A^{\prime})^{-1}}\right)^{T}")]{\kern
0.1pt\mathchar 866\relax\kern
0.1pt}{\rule{0.0pt}{650.43pt}}}{}}{2.4ex}}\stackon[-6.9pt]{\left({A(A^{\prime})^{-1}}\right)^{T}}{\tmpbox}$,
where
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[width("\left({A(A^{\prime})^{-1}}\right)^{T}")]{\kern
0.1pt\mathchar 866\relax\kern
0.1pt}{\rule{0.0pt}{650.43pt}}}{}}{2.4ex}}\stackon[-6.9pt]{\left({A(A^{\prime})^{-1}}\right)^{T}}{\tmpbox}$
is the projective automorphism corresponding to the matrix
$\left({A(A^{\prime})^{-1}}\right)^{T}$ as in (2.4). (See Lemma 4.28 in [6].)

######  Remark 6.16.

As $\gamma\searrow 1$ the $M_{\gamma}$ tend to the hypersurface $M_{1}$ which
is ${\mathbb{C}}$-convex but not strongly ${\mathbb{C}}$-convex. The matrix
$A_{1}$ is not invertible, but using the matrix

$A_{*}:=\begin{pmatrix}0&0&i\\\ 0&\gamma&0\\\ i&0&0\end{pmatrix}$

instead we obtain $w_{1}^{A_{*}}(\zeta)=\overline{\zeta}_{1}/|\zeta_{1}|$ and
$w_{2}^{A_{*}}(\zeta)=i|\zeta_{1}|-\zeta_{2}=-\operatorname{Re}\zeta_{2}$.
Thus $M_{1}$ is mapped onto the 2-manifold $S^{1}\times{\mathbb{R}}$.
$\lozenge$

###  6.2. Distinguished measures on $M_{\gamma}$

In Section 2.4.1 we previewed several measures now to be discussed in detail.
Each of these belongs to one of the families

${\mathscr{F}}_{r}=\\{c\mu_{r}:c>0\\},$

where $\mu_{r}=\alpha^{r}\,ds\wedge d\alpha\wedge d\theta$, so Leray
boundedness results obtained in previous sections are immediately applicable.

####  6.2.1. The pairing measure

Let $A\in GL(3,{\mathbb{C}})$ be as in (6.3) and ${\mathcal{S}}$ an
$A$-admissible, strongly ${\mathbb{C}}$-convex hypersurface in
${\mathbb{C}}^{2}$. Define the pairing measure (with respect to $A$) on
${\mathcal{S}}$ by the following formula:

(6.17) $\nu^{A}:=\frac{1}{(2\pi
i)^{2}}\Bigg{(}\left(dw_{1,A},dw_{2,A}\right)\begin{pmatrix}m_{11}&m_{12}\\\
m_{21}&m_{22}\end{pmatrix}\begin{pmatrix}a_{2}\\\ -a_{1}\end{pmatrix}\\\
+\begin{vmatrix}m_{11}&m_{12}\\\
m_{21}&m_{22}\end{vmatrix}\left(w_{2,A}\,dw_{1,A}-w_{1,A}\,dw_{2,A}\right)\Bigg{)}\wedge
d\zeta_{1}\wedge d\zeta_{2}.$

The authors previously introduced this measure in [6]. (Related objects also
appear in Section 7 of [4] and Section 3.2 of [1].) This measure allows for
the Leray transform to be written in terms of the dual variables. We restate
Proposition 4.30 from [6], which gives a universal formula for the Leray
transform of any $A$-admissible ${\mathcal{S}}$:

######  Proposition 6.18.

Let $A\in GL(3,{\mathbb{C}})$ be as in (6.3) and ${\mathcal{S}}$ an
$A$-admissible, strongly ${\mathbb{C}}$-convex hypersurface in
${\mathbb{C}}^{2}$. Then the Leray integral from (2.1a) may be written as

(6.19)
$\bm{L}_{{\mathcal{S}}}f(z)=\int_{\zeta\in{\mathcal{S}}}\frac{f(\zeta)}{\big{[}\Phi_{A}(z,w^{A}(\zeta))\big{]}^{2}}\nu^{A}(\zeta),$

where $\Phi_{A}$ is defined in (6.1).

In the special case of ${\mathcal{S}}=M_{\gamma}$, $A=A_{\gamma}$ we have

$\displaystyle\nu^{A_{\gamma}}=-\frac{i\gamma}{4\pi^{2}}\,dw_{1}^{A_{\gamma}}\wedge
d\zeta_{1}\wedge d\zeta_{2}$
$\displaystyle=\frac{\gamma^{2}}{8\pi^{2}i}|\zeta_{1}|^{\gamma-2}\,d\zeta_{2}\wedge
d\overline{\zeta}_{1}\wedge d\zeta_{1}$ (6.20)
$\displaystyle=\frac{\gamma^{2}}{4\pi^{2}}\alpha^{\gamma-1}ds\wedge
d\alpha\wedge d\theta.$

####  6.2.2. The preferred measure and its dual

In Section 2.2.3, we defined (up to a constant) the preferred measure
$\widetilde{\mu}_{{\mathcal{S}}}$. This measure was introduced by the first
author in [4] to define a pair of dual Hardy spaces with desirable projective
transformation laws. Employing the Fefferman measure (2.8) to construct such
spaces still leads to such transformation laws, but the preferred measure
$\widetilde{\mu}_{\mathcal{S}}$ dovetails with a pairing of these spaces in
the spirit of (6.19) in a way that $\mu_{\mathcal{S}}^{\sf Fef}$ does not.

One Hardy space is defined using $\widetilde{\mu}_{{\mathcal{S}}}$. To define
the other Hardy space, introduce the preferred dual measure by pulling back
the preferred measure on the dual ${\mathcal{S}}^{*,A}$ via the map $w^{A}$:

(6.21)
$\widetilde{\mu}_{{\mathcal{S}}}^{*,A}:=\left(w^{A}\right)^{*}\left(\widetilde{\mu}_{{\mathcal{S}}^{*,A}}\right).$

When the hypersurface and matrix are clear from context, we may simply write
$\widetilde{\mu}$ and $\widetilde{\mu}^{*}$ to denote the preferred measure
and its dual.

The preferred measure on $M_{\gamma}$ was already recorded in (2.13):

$\widetilde{\mu}_{M_{\gamma}}=c_{2}\frac{\gamma^{4/3}}{8(\gamma-1)^{1/3}i}\,|\zeta_{1}|^{\frac{\gamma-2}{3}}\,d\zeta_{2}\wedge
d\overline{\zeta}_{1}\wedge d\zeta_{1}.$

We will chose a convenient normalization constant $c_{2}$ shortly. But first
let us compute the dual measure. From (6.11) we obtain

$d\overline{w}_{1}^{A_{\gamma}}\wedge
dw_{1}^{A_{\gamma}}=(1-\gamma)|\zeta_{1}|^{2\gamma-4}\,d\overline{\zeta}_{1}\wedge
d\zeta_{1},$

so from (6.21), we see

$\displaystyle\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}}$
$\displaystyle=\frac{c_{2}\cdot\left(\gamma^{*}\right)^{4/3}|w_{1}^{A_{\gamma}}|^{\frac{\gamma^{*}-2}{3}}}{8(\gamma^{*}-1)^{1/3}i}\,dw_{2}^{A_{\gamma}}\wedge
d\overline{w}_{1}^{A_{\gamma}}\wedge dw_{1}^{A_{\gamma}}$ (6.22)
$\displaystyle=\frac{c_{2}\cdot\gamma^{4/3}}{8(\gamma-1)i}|\zeta_{1}|^{\frac{5(\gamma-2)}{3}}\,d\zeta_{2}\wedge
d\overline{\zeta}_{1}\wedge d\zeta_{1}.$

Note the Radon-Nikodym derivative

(6.23) $\frac{{\rm d}\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}}}{{\rm
d}\widetilde{\mu}_{M_{\gamma}}}=\frac{|\zeta_{1}|^{\frac{4}{3}(\gamma-2)}}{(\gamma-1)^{2/3}}.$

Now define the geometric mean
$\sqrt{\widetilde{\mu}_{M_{\gamma}}\,\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}}}$
of measures $\widetilde{\mu}_{M_{\gamma}}$ and
$\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}}$ by

${\rm
d}\sqrt{\widetilde{\mu}_{M_{\gamma}}\,\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}}}:=\sqrt{\frac{{\rm
d}\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}}}{{\rm
d}\widetilde{\mu}_{M_{\gamma}}}}\>{\rm d}\widetilde{\mu}_{M_{\gamma}}.$

Therefore

$\sqrt{\widetilde{\mu}_{M_{\gamma}}\,\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}}}=\dfrac{c_{2}\gamma^{4/3}|\zeta_{1}|^{\gamma-2}}{8(\gamma-1)^{2/3}i}\,d\zeta_{2}\wedge
d\overline{\zeta}_{1}\wedge d\zeta_{1}$

and thus,

$\displaystyle\left|\int_{M_{\gamma}}fg\,{\rm d}\nu^{A_{\gamma}}\right|$
$\displaystyle\leq\sqrt{\int_{M_{\gamma}}|f|^{2}\,{\rm
d}\widetilde{\mu}_{M_{\gamma}}}\sqrt{\int_{M_{\gamma}}|g|^{2}\,{\rm
d}\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}}}\frac{{\rm
d}\nu^{A_{\gamma}}}{{\rm
d}\sqrt{\widetilde{\mu}_{M_{\gamma}}\,\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}}}}$
$\displaystyle=\frac{\left(\gamma(\gamma-1)\right)^{2/3}}{c_{2}\pi^{2}}\|f\|^{2}_{L^{2}\left(M_{\gamma},\widetilde{\mu}\right)}\|g\|^{2}_{L^{2}\left(M_{\gamma},\widetilde{\mu}^{*}\right)}.$

(Here we identify the positive three-form $\nu^{A_{\gamma}}$ with a measure so
that $\int fg\,{\rm d}\nu^{A_{\gamma}}$ and $\int fg\,\nu^{A_{\gamma}}$ have
the same meaning.)

In order to obtain a sharp Cauchy-Schwarz inequality

(6.24)
$\left|\int_{M_{\gamma}}fg\,\nu^{A_{\gamma}}\right|\leq\|f\|^{2}_{L^{2}\left(M_{\gamma},\widetilde{\mu}\right)}\|g\|^{2}_{L^{2}\left(M_{\gamma},\widetilde{\mu}^{*}\right)},$

we set

$c_{2}=\frac{\left(\gamma(\gamma-1)\right)^{2/3}}{\pi^{2}}$

so that

(6.25)
$\sqrt{\widetilde{\mu}_{M_{\gamma}}\,\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}}}=\dfrac{\gamma^{2}|\zeta_{1}|^{\gamma-2}}{8\pi^{2}i}\,d\zeta_{2}\wedge
d\overline{\zeta}_{1}\wedge d\zeta_{1}=\nu^{A_{\gamma}}.$

For each $f$, there is some non-zero $g$ such that equality holds in (6.24).
(The same is true with the roles of $f$ and $g$ reversed.) Here, we’ve allowed
$c_{2}$ to depend not only on the dimension but also on the choice of matrix
$A_{\gamma}$.

This choice of $c_{2}$ will be fixed throughout the rest of the paper. We
therefore settle on the convention that

$\displaystyle\widetilde{\mu}_{M_{\gamma}}$
$\displaystyle=\frac{\gamma^{2}(\gamma-1)^{1/3}}{8\pi^{2}i}|\zeta_{1}|^{\frac{\gamma-2}{3}}\,d\zeta_{2}\wedge
d\overline{\zeta}_{1}\wedge d\zeta_{1}$ (6.26)
$\displaystyle=\frac{\gamma^{2}(\gamma-1)^{1/3}}{4\pi^{2}}\alpha^{\frac{\gamma+1}{3}}ds\wedge
d\alpha\wedge d\theta$

and

$\displaystyle\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}}$
$\displaystyle=\frac{\gamma^{2}}{8\pi^{2}(\gamma-1)^{1/3}i}|\zeta_{1}|^{\frac{5\gamma-10}{3}}\,d\zeta_{2}\wedge
d\overline{\zeta}_{1}\wedge d\zeta_{1}$ (6.27)
$\displaystyle=\frac{\gamma^{2}}{4\pi^{2}(\gamma-1)^{1/3}}\alpha^{\frac{5\gamma-7}{3}}ds\wedge
d\alpha\wedge d\theta.$

###  6.3. Dual Hardy spaces and the dual Leray transform

Formulas (6.2.1), (6.2.2) and (6.2.2) show that each of the distinguished
measures from the previous section are contained in some ${\mathscr{F}}_{r}$.
In particular,

(6.28)
$\nu^{A_{\gamma}}\in{\mathscr{F}}_{\gamma-1},\qquad\widetilde{\mu}_{M_{\gamma}}\in{\mathscr{F}}_{\frac{\gamma+1}{3}},\qquad\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}}\in{\mathscr{F}}_{\frac{5\gamma-7}{3}}.$

Notice that for any $\gamma>1$, the numbers
$\gamma-1,\tfrac{\gamma+1}{3},\tfrac{5\gamma-7}{3}\in{\mathcal{I}}_{0}=(-1,2\gamma-1)$.
Therefore, Corollary 5.28 implies that the Leray transform is a bounded
operator on all three spaces $L^{2}(M_{\gamma},\nu^{A_{\gamma}})$,
$L^{2}(M_{\gamma},\widetilde{\mu}_{M_{\gamma}})$ and
$L^{2}(M_{\gamma},\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}})$.

In what follows, $\left(w^{A_{\gamma}}\right)^{*}$ and
$\left(w^{A_{\gamma}}\right)_{*}$ refer to the pullback and pushforward of
$w^{A_{\gamma}}$, respectively. In the spirit of (5.49), define the Hardy
space

$H^{2}(M_{\gamma},\widetilde{\mu}_{M_{\gamma}}):=\bm{L}_{\gamma}\big{(}L^{2}(M_{\gamma},\widetilde{\mu}_{M_{\gamma}})\big{)}=\ker({{\bm{L}}_{\gamma}}-I)$

and the dual Hardy space

(6.29) $H^{2}_{\sf
dual}\big{(}M_{\gamma},\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}}\big{)}:=\left(w^{A_{\gamma}}\right)^{*}\Big{(}H^{2}\big{(}M_{\gamma^{*}},\widetilde{\mu}_{{M_{\gamma^{*}}}}\big{)}\Big{)}.$

Also define the dual Leray transform

(6.30)
$\widetilde{\bm{L}}_{\gamma}:=\left(w^{A_{\gamma}}\right)^{*}\circ{\bm{L}}_{\gamma^{*}}\circ\left(w^{A_{\gamma}}\right)_{*}.$

######  Proposition 6.31.

The dual Leray transform $\widetilde{\bm{L}}_{\gamma}$ is a bounded projection
operator from
$L^{2}\big{(}M_{\gamma},\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}}\big{)}$
onto $H^{2}_{\sf
dual}\big{(}M_{\gamma},\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}}\big{)}$.

###### Proof.

We occasionally write $\widetilde{\mu}^{*}$ for
$\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}}$ when context is clear. Letting
$g\in L^{2}(M_{\gamma},\widetilde{\mu}^{*})$ and writing the purely imaginary
constant in (6.2.2) as $\frac{\gamma^{2}}{8\pi^{2}(\gamma-1)^{1/3}i}:=-Ci$, we
have

$\displaystyle\left\|g\right\|^{2}_{\widetilde{\mu}^{*}}$
$\displaystyle=-Ci\int_{M_{\gamma}}|g(\zeta)|^{2}|\zeta_{1}|^{\frac{5\gamma-10}{3}}\,d\zeta_{2}\wedge
d\overline{\zeta}_{1}\wedge d\zeta_{1}$ (6.32)
$\displaystyle=-Ci\int_{M_{\gamma^{*}}}|g\circ
w^{-1}|^{2}|w_{1}|^{\frac{2-\gamma}{3\gamma-3}}\,dw_{2}\wedge
d\overline{w}_{1}\wedge dw_{1}<\infty.$

Since $-Ci\,|w_{1}|^{\frac{2-\gamma}{3\gamma-3}}\,dw_{2}\wedge
d\overline{w}_{1}\wedge dw_{1}\in{\mathcal{F}}_{\frac{2\gamma-1}{3\gamma-3}}$,
we see $g\circ w^{-1}\in
L^{2}\Big{(}M_{\gamma^{*}},\mu_{\frac{2\gamma-1}{3\gamma-3}}\Big{)}$ and that
each function in this space corresponds to a unique such $g$. Now considering
that

$\tfrac{2\gamma-1}{3\gamma-3}\in(-1,2\gamma^{*}-1)={\mathcal{I}}_{0}$

(the interval of $r$ values with Leray boundedness on
$L^{2}(M_{\gamma^{*}},\mu_{r})$), Theorem 1.4 says that $\bm{L}_{\gamma^{*}}$
is bounded on
$L^{2}\Big{(}M_{\gamma^{*}},\mu_{\frac{2\gamma-1}{3\gamma-3}}\Big{)}$. Using
$w^{A_{\gamma}}=w$ to pull the computation back to $M_{\gamma}$, (since
$\widetilde{\bm{L}}_{\gamma}g=w^{*}\circ\bm{L}_{\gamma^{*}}\circ w_{*}\circ
g$) we conclude that
$\left\|\widetilde{\bm{L}}_{\gamma}g\right\|_{\widetilde{\mu}^{*}}\lesssim\left\|g\right\|_{\widetilde{\mu}^{*}}$,
where the implied constant is independent of the particular choice of $g$. We
are also now able to invoke Proposition 5.48 to conclude that the dual Leray
transform is a projection.

Since
$\widetilde{\mu}_{M_{\gamma^{*}}}\in{\mathcal{F}}_{\frac{2\gamma-1}{3\gamma-3}}$,
$H^{2}\big{(}M_{\gamma^{*}},\widetilde{\mu}_{{M_{\gamma^{*}}}}\big{)}$ is the
image of $L^{2}\Big{(}M_{\gamma^{*}},\mu_{\frac{2\gamma-1}{3\gamma-3}}\Big{)}$
under $\bm{L}_{\gamma^{*}}$. Definition (6.29) now lets us conclude

$\widetilde{\bm{L}}_{\gamma}\Big{(}L^{2}\big{(}M_{\gamma},\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}}\big{)}\Big{)}=H^{2}_{\sf
dual}\big{(}M_{\gamma},\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}}\big{)}.$

∎

######  Remark 6.33.

It is shown in the Appendix that functions in the Hardy space
$H^{2}(M_{\gamma},\widetilde{\mu}_{M_{\gamma}})$ are boundary values of
holomorphic functions on $\Omega_{\gamma}$. Functions in $H^{2}_{\sf
dual}\big{(}M_{\gamma},\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}}\big{)}$ are
pullbacks via $w^{A_{\gamma}}$ of holomorphic boundary values on
$\Omega_{\gamma^{*}}$. $\lozenge$

######  Remark 6.34.

If $Y$ is a non-vanishing vector field annihilating CR functions on
$M_{\gamma^{*}}$, then $H^{2}_{\sf
dual}\big{(}M_{\gamma},\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}}\big{)}$ can
be equivalently be thought of as the set of functions in
$L^{2}\big{(}M_{\gamma},\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}}\big{)}$
annihilated by $\left(w^{A_{\gamma}}\right)^{*}Y$. (So $Y$ induces the
“projective dual CR structure on $M_{\gamma}$”. See Section 3 of [7] for a
general discussion.) Alternatively, Lemma 4.40 in [6] leads to the
characterizing vector field of the dual Hardy space. It can be shown that

(6.35) $\overline{L}_{\sf
dual}:=\frac{i}{2}\frac{\partial}{\partial\zeta_{1}}+\frac{(2-\gamma)i}{2\gamma}\frac{\overline{\zeta}_{1}}{\zeta_{1}}\frac{\partial}{\partial\overline{\zeta}_{1}}-\frac{\gamma}{2}\,\overline{\zeta}_{1}|\zeta_{1}|^{\gamma-2}\frac{\partial}{\partial\zeta_{2}}+\left(1-\frac{\gamma}{2}\right)\overline{\zeta}_{1}|\zeta_{1}|^{\gamma-2}\frac{\partial}{\partial\overline{\zeta}_{2}}$

annihilates all functions in $H^{2}_{\sf
dual}\big{(}M_{\gamma},\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}}\big{)}$.
$\lozenge$

$\widetilde{\bm{L}}_{\gamma}$ is now re-expressed as a single integral:

######  Proposition 6.36.

Let $g\in
L^{2}\big{(}M_{\gamma},\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}}\big{)}$.
The dual Leray transform is obtained by calculating the following integral

(6.37)
$\widetilde{\bm{L}}_{\gamma}(g)(z)=\frac{\gamma^{2}}{8\pi^{2}i}\int_{M_{\gamma}}g(\zeta)\frac{|\zeta_{1}|^{\gamma-2}d\zeta_{2}\wedge
d\overline{\zeta}_{1}\wedge
d\zeta_{1}}{\left[\gamma\overline{z}_{1}|z_{1}|^{\gamma-2}(z_{1}-\zeta_{1})+i(z_{2}-\zeta_{2})\right]^{2}}.$

###### Proof.

The map $w^{A_{\gamma}}=w$ given by (6.12) is a diffeomorphism from
$M_{\gamma}$ onto $M_{\gamma^{*}}$. To help keep track of notation, write

$\displaystyle\zeta$ $\displaystyle=(\zeta_{1},\zeta_{2})\in M_{\gamma},\qquad
w(\zeta):=\zeta^{*}=(\zeta_{1}^{*},\zeta_{2}^{*})\in M_{\gamma^{*}},$
$\displaystyle z$ $\displaystyle=(z_{1},z_{2})\in M_{\gamma},\qquad
w(z):=z^{*}=(z_{1}^{*},z_{2}^{*})\in M_{\gamma^{*}}.$

Now transfer the computation from $M_{\gamma}$ to $M_{\gamma^{*}}$ by
considering the pushforward of $g$, denoted $G:=w_{*}(g)$. The proof of
Proposition 6.31 shows both that $G\in
L^{2}\Big{(}M_{\gamma^{*}},\mu_{\frac{2\gamma-1}{3\gamma-3}}\Big{)}$ and that
$\bm{L}_{\gamma^{*}}$ is a bounded operator on this space.

(6.38)
$\displaystyle\widetilde{\bm{L}}_{\gamma}(g)(z)=(w^{*}\circ\bm{L}_{\gamma^{*}}\circ
w_{*}\circ g)(z)$ $\displaystyle=\left(\bm{L}_{\gamma^{*}}(g\circ
w^{-1})\right)(w(z))=\bm{L}_{\gamma^{*}}(G)(z^{*}).$

Now,

$\displaystyle\bm{L}_{\gamma^{*}}(G)(z^{*})$
$\displaystyle=\frac{(\gamma^{*})^{2}}{8\pi^{2}i}\int_{M_{\gamma^{*}}}G(\zeta^{*})\frac{|\zeta_{1}^{*}|^{{\gamma^{*}}-2}d\zeta^{*}_{2}\wedge
d\overline{\zeta^{*}_{1}}\wedge
d\zeta^{*}_{1}}{\left[\gamma^{*}\overline{\zeta_{1}^{*}}|\zeta_{1}^{*}|^{\gamma^{*}-2}(\zeta_{1}^{*}-z_{1}^{*})+i(\zeta_{2}^{*}-z_{2}^{*})\right]^{2}}$
$\displaystyle=\frac{\gamma^{2}}{8\pi^{2}(\gamma-1)^{2}i}\int_{M_{\gamma^{*}}}G(\zeta^{*})\frac{|\zeta_{1}^{*}|^{\frac{2-\gamma}{\gamma-1}}d\zeta^{*}_{2}\wedge
d\overline{\zeta^{*}_{1}}\wedge
d\zeta^{*}_{1}}{\left[(\frac{\gamma}{\gamma-1})\overline{\zeta^{*}_{1}}|\zeta_{1}^{*}|^{\frac{2-\gamma}{\gamma-1}}(\zeta_{1}^{*}-z_{1}^{*})+i(\zeta_{2}^{*}-z_{2}^{*})\right]^{2}}$
(6.39)
$\displaystyle=\frac{\gamma^{2}}{8\pi^{2}i}\int_{M_{\gamma^{*}}}G(\zeta^{*})\frac{|\zeta_{1}^{*}|^{\frac{2-\gamma}{\gamma-1}}d\zeta^{*}_{2}\wedge
d\overline{\zeta^{*}_{1}}\wedge
d\zeta^{*}_{1}}{\left[\gamma|\zeta_{1}^{*}|^{\frac{\gamma}{\gamma-1}}-\gamma
z_{1}^{*}\overline{\zeta^{*}_{1}}|\zeta_{1}^{*}|^{\frac{2-\gamma}{\gamma-1}}+i(\gamma-1)(\zeta_{2}^{*}-z_{2}^{*})\right]^{2}}.$

Now we wish to move the computation back to $M_{\gamma}$. Writing (6.11) and
(6.13) in the notational convention adopted above we see that

(6.40) $\displaystyle\zeta_{1}^{*}$
$\displaystyle=\overline{\zeta_{1}}|\zeta_{1}|^{\gamma-2},\qquad\zeta_{2}^{*}=\frac{\zeta_{2}-i\gamma|\zeta_{1}|^{\gamma}}{1-\gamma},$
(6.41) $\displaystyle\zeta_{1}$
$\displaystyle=\overline{\zeta_{1}^{*}}|\zeta_{1}^{*}|^{\frac{2-\gamma}{\gamma-1}},\qquad\zeta_{2}=(1-\gamma)\zeta_{2}^{*}+i\gamma|\zeta_{1}|^{\frac{\gamma}{\gamma-1}}.$

The same relations also hold between the $z$ and $z^{*}$ variables. From here,
it follows that

(6.42) $d\zeta^{*}_{2}\wedge d\overline{\zeta^{*}_{1}}\wedge
d\zeta^{*}_{1}=|\zeta_{1}|^{2\gamma-4}d\zeta_{2}\wedge
d\overline{\zeta_{1}}\wedge d\zeta_{1}.$

Using (6.40) and (6.41), rewrite the term inside the brackets appearing in the
denominator in (6.39):

$\displaystyle\gamma|\zeta_{1}^{*}|^{\frac{\gamma}{\gamma-1}}-\gamma
z_{1}^{*}\overline{\zeta^{*}_{1}}|\zeta_{1}^{*}|^{\frac{2-\gamma}{\gamma-1}}+i(\gamma-1)(\zeta_{2}^{*}-z_{2}^{*})$
$\displaystyle=\,\,$
$\displaystyle\gamma|\zeta_{1}|^{\gamma}-\gamma\zeta_{1}\overline{z_{1}}|z_{1}|^{\gamma-2}+\gamma|z_{1}|^{\gamma}-\gamma|\zeta_{1}|^{\gamma}+i(z_{2}-\zeta_{2})$
(6.43) $\displaystyle=\,\,$
$\displaystyle\gamma\overline{z_{1}}|z_{1}|^{\gamma-2}(z_{1}-\zeta_{1})+i(z_{2}-\zeta_{2})$

Inserting equations (6.42) and (6.43) into (6.39) shows

$\eqref{E:integral-
in-*variables}=\frac{\gamma^{2}}{8\pi^{2}i}\int_{M_{\gamma}}g(\zeta)\frac{|\zeta_{1}|^{\gamma-2}d\zeta_{2}\wedge
d\overline{\zeta}_{1}\wedge
d\zeta_{1}}{\left[\gamma\overline{z}_{1}|z_{1}|^{\gamma-2}(z_{1}-\zeta_{1})+i(z_{2}-\zeta_{2})\right]^{2}},$

which is equal to $\widetilde{\bm{L}}_{\gamma}(g)(z)$ by (6.38). ∎

The preceding proposition demonstrates the close connection between the dual
Leray transform and the adjoint of the Leray transform with respect to the
measure $\sigma$:

######  Corollary 6.44.

The dual Leray transform on $M_{\gamma}$ can be expressed as

$\displaystyle\widetilde{\bm{L}}_{\gamma}=\mathfrak{c}\circ\bm{L}^{(*,\sigma)}_{\gamma}\circ\mathfrak{c},$

where $\mathfrak{c}$ is the conjugation operator and $\bm{L}^{(*,\sigma)}$ is
given by the integral in (5.37).

###### Proof.

This is immediate from formulas (5.37) and (6.37). ∎

######  Remark 6.45.

We’ve seen that $\bm{L}_{\gamma}$ is bounded on $L^{2}(M_{\gamma},\mu_{r})$ if
and only if $r\in{\mathcal{I}}_{0}=(-1,2\gamma-1)$. For a fixed $r$ in this
range, choose $r^{\prime}$ with $\frac{r+r^{\prime}}{2}=\gamma-1$, and let
$f\in L^{2}(M_{\gamma},\mu_{r})$, $g\in L^{2}(M_{\gamma},\mu_{r^{\prime}})$.
By symmetry (Remark 5.29), $r^{\prime}$ is also contained in
${\mathcal{I}}_{0}$. General theory (or, alternatively, the proof of Theorem
5.41) shows

$\langle\bm{L}_{\gamma}f,g\rangle_{\sigma}=\langle
f,\bm{L}_{\gamma}^{(*,\sigma)}g\rangle_{\sigma},$

implying that

$\left\|\bm{L}_{\gamma}\right\|_{\mu_{r}}=\|\bm{L}_{\gamma}^{(*,\sigma)}\|_{\mu_{r^{\prime}}}=\|\widetilde{\bm{L}}_{\gamma}\|_{\mu_{r^{\prime}}}.$

This shows that the dual Leray transform $\widetilde{\bm{L}}_{\gamma}$ is
bounded on $L^{2}(M_{\gamma},\mu_{r^{\prime}})$ for all
$r^{\prime}\in{\mathcal{I}}_{0}$. In particular,

$\left\|\bm{L}_{\gamma}\right\|_{\widetilde{\mu}}=\|\widetilde{\bm{L}}_{\gamma}\|_{\widetilde{\mu}^{*}}.$

where $\widetilde{\mu}$ and $\widetilde{\mu}^{*}$ are the preferred measure
and preferred dual measure, respectively. $\lozenge$

###  6.4. Pairing Hardy spaces

For $f\in L^{2}\left(M_{\gamma},\widetilde{\mu}_{M_{\gamma}}\right)$, $g\in
L^{2}\big{(}M_{\gamma},\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}}\big{)}$,
define the bilinear pairing

(6.46) $\langle\\!\langle
f,g\rangle\\!\rangle=\int_{M_{\gamma}}f(\zeta)g(\zeta)\,\nu^{A_{\gamma}}(\zeta).$

Continue to write $\widetilde{\mu}$ and $\widetilde{\mu}^{*}$ for
$\widetilde{\mu}_{M_{\gamma}}$ and
$\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}}$, respectively. This pairing
facilitates the representation of linear functionals on
$H^{2}(M_{\gamma},\widetilde{\mu})$ by functions in $H^{2}_{\sf
dual}\left(M_{\gamma},\widetilde{\mu}^{*}\right)$. In Theorem 6.52 below we
prove that the efficiency of this representation is closely connected to the
Leray transform. We point out that:

  * (1)

The preferred measure $\widetilde{\mu}$ and preferred dual measure
$\widetilde{\mu}^{*}$ are projectively invariant, so the corresponding Hardy
spaces are too. Compare this to both Theorem 2 and Section 8 in [4].

  * (2)

Pairing $H^{2}(M_{\gamma},\widetilde{\mu})$ and $H^{2}_{\sf
dual}\left(M_{\gamma},\widetilde{\mu}^{*}\right)$ via (6.46) is closely tied
to the universal expression of the Leray transform given in (6.19).

######  Lemma 6.47.

For $f\in L^{2}\left(M_{\gamma},\widetilde{\mu}\right)$, $g\in
L^{2}\big{(}M_{\gamma},\widetilde{\mu}^{*}\big{)}$ we have

(6.48) $\langle\\!\langle\bm{L}_{\gamma}f,g\rangle\\!\rangle=\langle\\!\langle
f,\widetilde{\bm{L}}_{\gamma}g\rangle\\!\rangle=\langle\\!\langle\bm{L}_{\gamma}f,\widetilde{\bm{L}}_{\gamma}g\rangle\\!\rangle.$

###### Proof.

Since $\bm{L}_{\gamma}$ and $\widetilde{\bm{L}}_{\gamma}$ are projections, it
suffices to prove the first equality. Notice that

$\displaystyle\langle\\!\langle\bm{L}_{\gamma}f,g\rangle\\!\rangle=\langle\bm{L}_{\gamma}f,\overline{g}\rangle_{\nu}=\langle
f,\bm{L}_{\gamma}^{(*\sigma)}\overline{g}\rangle_{\nu}=\langle\\!\langle
f,\mathfrak{c}\circ\bm{L}^{(*,\sigma)}_{\gamma}\circ\mathfrak{c}(g)\rangle\\!\rangle=\langle\\!\langle
f,\widetilde{\bm{L}}_{\gamma}g\rangle\\!\rangle.$

The second equality is justified by the proof of Theorem 5.41 and the final
equality is just Corollary 6.44. ∎

######  Remark 6.49.

Formula (6.48) could also be proved by adapting the Plemelj-formula-based
argument found in Theorem 25 of [4] to the current unbounded setting.
$\lozenge$

Now consider the map

$\displaystyle\chi_{\gamma}\colon
L^{2}\big{(}M_{\gamma},\widetilde{\mu}^{*}\big{)}\to\left(L^{2}\left(M_{\gamma},\widetilde{\mu}\right)\right)^{*}$

given by

$\displaystyle\chi_{\gamma}(g)\colon f\to\langle\\!\langle
f,g\rangle\\!\rangle,$

along with the companion map

$\displaystyle\widehat{\chi}_{\gamma}\colon H^{2}_{\sf
dual}\big{(}M_{\gamma},\widetilde{\mu}^{*}\big{)}\to\left(H^{2}\big{(}M_{\gamma},\widetilde{\mu}\big{)}\right)^{*}$

given by

$\displaystyle\widehat{\chi}_{\gamma}(g)=\chi_{\gamma}(g)\Big{|}_{H^{2}(M_{\gamma},\widetilde{\mu})}.$

Note that $\chi_{\gamma}$ and $\widehat{\chi}_{\gamma}$ are linear (not
conjugate-linear). From (6.24) we see that $\chi_{\gamma}$ is an isometry and
that

$\displaystyle\left\|\widehat{\chi}_{\gamma}\right\|$ $\displaystyle\leq 1.$

Define a map
$\widetilde{\bm{R}}:L^{2}\big{(}M_{\gamma},\widetilde{\mu}^{*}\big{)}\to
L^{2}\left(M_{\gamma},\widetilde{\mu}\right)$ by

(6.50) $f(\zeta)\longmapsto\sqrt{\frac{{\rm
d}\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}}}{{\rm
d}\widetilde{\mu}_{M_{\gamma}}}}f(\zeta)=\frac{|\zeta_{1}|^{\frac{2}{3}(\gamma-2)}}{(\gamma-1)^{1/3}}f(\zeta):=\widetilde{\bm{R}}f,$

where the Radon-Nikodym derivative is calculated in (6.23).

######  Lemma 6.51.

The map
$\widetilde{\bm{R}}:L^{2}\big{(}M_{\gamma},\widetilde{\mu}_{M_{\gamma}}^{*,A_{\gamma}}\big{)}\to
L^{2}\left(M_{\gamma},\widetilde{\mu}_{M_{\gamma}}\right)$ is an isometry.

###### Proof.

The appearance of the Radon-Nikodym derivative makes this clear. ∎

######  Theorem 6.52.

The operator $\widehat{\chi}_{\gamma}\colon H^{2}_{\sf
dual}\big{(}M_{\gamma},\widetilde{\mu}^{*}\big{)}\to\left(H^{2}\big{(}M_{\gamma},\widetilde{\mu}\big{)}\right)^{*}$
is invertible with norm

(6.53)
$\left\|{\widehat{\chi}}_{\gamma}^{\,-1}\right\|=\left\|\bm{L}_{\gamma}\right\|_{\widetilde{\mu}}.$

######  Remark 6.54.

If $M_{\gamma}$ were the boundary of a smoothly bounded strongly
${\mathbb{C}}$-convex domain in ${\mathbb{C}}^{2}$ (or
${\mathbb{C}}{\mathbb{P}}^{2}$), then Theorem 6.52 would just be a paraphrase
of Corollary 26 from [4]; the argument offered below is an adaptation of that
earlier proof. $\lozenge$

###### Proof.

Let $g\in\ker\widehat{\chi}_{\gamma}$, i.e., $\widehat{\chi}_{\gamma}(g)$ is
the zero functional on $H^{2}\big{(}M_{\gamma},\widetilde{\mu}\big{)}$. It can
be easily checked that $\bm{L}_{\gamma}\widetilde{\bm{R}}\,\overline{g}\in
H^{2}\big{(}M_{\gamma},\widetilde{\mu}\big{)}$. This means that

$\displaystyle
0=\widehat{\chi}_{\gamma}(g)\left(\bm{L}_{\gamma}\widetilde{\bm{R}}\,\overline{g}\right)=\left\langle\\!\left\langle\bm{L}_{\gamma}\widetilde{\bm{R}}\,\overline{g},g\right\rangle\\!\right\rangle=\left\langle\\!\left\langle\widetilde{\bm{R}}\,\overline{g},\widetilde{\bm{L}}_{\gamma}g\right\rangle\\!\right\rangle=\left\langle\\!\left\langle\widetilde{\bm{R}}\,\overline{g},g\right\rangle\\!\right\rangle=\left\|g\right\|_{\widetilde{\mu}^{*}}^{2}.$

Thus $g\equiv 0$ and $\widehat{\chi}_{\gamma}$ is injective.

Let $T\in\left(H^{2}\big{(}M_{\gamma},\widetilde{\mu}\big{)}\right)^{*}$.
There is some $h\in L^{2}\big{(}M_{\gamma},\widetilde{\mu}^{*}\big{)}$
representing $T$ and satisfying

$\|h\|_{L^{2}(M_{\gamma},\widetilde{\mu}^{*})}=\|T\|_{H^{2}(M_{\gamma},\widetilde{\mu})^{*}}.$

Now for all $f\in H^{2}(M_{\gamma},\widetilde{\mu})$,

$\displaystyle Tf$ $\displaystyle=\left\langle\\!\left\langle
f,h\right\rangle\\!\right\rangle=\left\langle\\!\left\langle\bm{L}_{\gamma}f,h\right\rangle\\!\right\rangle=\left\langle\\!\left\langle
f,\widetilde{\bm{L}}_{\gamma}h\right\rangle\\!\right\rangle=\widehat{\chi}_{\gamma}\left(\widetilde{\bm{L}}_{\gamma}h\right)(f)$

In other words,
$T=\widehat{\chi}_{\gamma}\left(\widetilde{\bm{L}}_{\gamma}h\right)$ with
$\|\widetilde{\bm{L}}_{\gamma}h\|_{\widetilde{\mu}^{*}}\leq\|\widetilde{\bm{L}}_{\gamma}\|_{\widetilde{\mu}^{*}}\|h\|_{\widetilde{\mu}^{*}}=\|{\bm{L}}_{\gamma}\|_{\widetilde{\mu}}\|T\|$.
So $\widehat{\chi}_{\gamma}$ is surjective with
$\|\widehat{\chi}_{\gamma}^{\,-1}\|\leq\|{\bm{L}}_{\gamma}\|_{\widetilde{\mu}}$.

To prove the reverse estimate let $f\in H^{2}(M_{\gamma},\widetilde{\mu})$ and
pick $h\in L^{2}\big{(}M_{\gamma},\widetilde{\mu}^{*}\big{)}$ such that
$\left\|h\right\|_{\widetilde{\mu}^{*}}=1$. Then

$\displaystyle\left|\widehat{\chi}_{\gamma}(\widetilde{\bm{L}}_{\gamma}h)(f)\right|=\left|\langle\\!\langle
f,\widetilde{\bm{L}}_{\gamma}h\rangle\\!\rangle\right|=\left|\langle\\!\langle\bm{L}_{\gamma}f,h\rangle\\!\rangle\right|=\left|\langle\\!\langle
f,h\rangle\\!\rangle\right|\leq\|f\|_{\widetilde{\mu}}$

and so $\|\widehat{\chi}_{\gamma}(\widetilde{\bm{L}}_{\gamma}h)\|\leq 1$ for
any such $h$. Let $\epsilon>0$ and choose a specific such $h$ so that
$\|\widetilde{\bm{L}}_{\gamma}h\|_{\widetilde{\mu}^{*}}\geq\|{\bm{L}}_{\gamma}\|_{\widetilde{\mu}}-\epsilon$.
It easily follows that
$\|\widehat{\chi}_{\gamma}^{-1}\|\geq\|\widetilde{\bm{L}}_{\gamma}h\|_{\widetilde{\mu}^{*}}\geq\|{\bm{L}}_{\gamma}\|_{\widetilde{\mu}}-\epsilon$.
Since $\epsilon$ was arbitrary we have
$\|\widehat{\chi}_{\gamma}^{-1}\|\geq\|{\bm{L}}_{\gamma}\|_{\widetilde{\mu}}$,
concluding the proof. ∎

See Section 9 in [4] for related results.

##  Appendix A Constructing holomorphic functions on $\Omega_{\gamma}$

We verify here that for certain values of $r$, the Leray transform maps
functions $f\in L^{2}(M_{\gamma},\mu_{r})$ to holomorphic functions on the
domain
$\Omega_{\gamma}=\left\\{(z_{1},z_{2})\in{\mathbb{C}}^{2}\colon\text{Im}(z_{2})>\left|z_{1}\right|^{\gamma}\right\\}.$

The main result is the following:

######  Theorem A.1.

Let $\bm{L}$ denote the Leray transform. $\bm{L}$ maps
$L^{2}(M_{\gamma},\mu_{r})\to{\mathcal{O}}(\Omega_{\gamma})$ for each
$r\in{\mathcal{J}}_{0}:=(-\gamma-1,2\gamma-1)$.

Notice that ${\mathcal{J}}_{0}\supsetneq{\mathcal{I}}_{0}=(-1,2\gamma-1)$, the
interval of $r$ values for which $\bm{L}$ is bounded from
$L^{2}(M_{\gamma},\mu_{r})\to L^{2}(M_{\gamma},\mu_{r})$. In Section 5.3, we
defined the Hardy space
$H^{2}(M_{\gamma},\mu_{r}):=\bm{L}\big{(}L^{2}(M_{\gamma},\mu_{r})\big{)}$ for
each $r\in{\mathcal{I}}_{0}$. Theorem A.1 confirms that these Hardy spaces can
be viewed as consisting entirely of boundary values of holomorphic functions.

###  A.1. Symmetries of the Leray kernel

Partition $\Omega_{\gamma}$ into translated copies of $M_{\gamma}$ by defining

(A.2)
$M_{\gamma}^{\epsilon}=\left\\{(z_{1},z_{2}):\text{Im}(z_{2})=|z_{1}|^{\gamma}+\epsilon\right\\}.$

It is clear that $\Omega_{\gamma}=\bigcup_{\epsilon>0}M_{\gamma}^{\epsilon}$
and that this is a union of disjoint sets.

Let $z\in\Omega_{\gamma}$. Equations (2.3a), (2.3b) and (2.3c) provide the
formula for the Leray transform:

(A.3)
$\bm{L}f(z)=\int_{M_{\gamma}}f(\zeta)\ell_{\rho}(z,\zeta)\lambda_{\rho}(\zeta)=\frac{\gamma^{2}}{{16\pi^{2}}}\int_{M_{\gamma}}f(\zeta)\ell_{\rho}(z,\zeta)\sigma(\zeta),$

where the kernel $\ell_{\rho}=\ell$ and measure $\sigma$ are

$\displaystyle\ell(z,\zeta)$
$\displaystyle=\frac{4}{\left(\gamma\overline{\zeta}_{1}|\zeta_{1}|^{\gamma-2}(\zeta_{1}-z_{1})+i(\zeta_{2}-z_{2})\right)^{2}},\qquad\sigma(\zeta)=\frac{1}{2i}|\zeta_{1}|^{\gamma-2}\,d\zeta_{2}\wedge
d\overline{\zeta}_{1}\wedge d\zeta_{1}.$

The measure $\sigma$ is special instance of $\mu_{r}$, given in rectangular
coordinates for $r\in{\mathbb{R}}$ by

$\displaystyle\mu_{r}(\zeta)=\frac{1}{2i}|\zeta_{1}|^{r-1}\,d\zeta_{2}\wedge
d\overline{\zeta}_{1}\wedge d\zeta_{1}.$

Recall the three types of automorphisms of $M_{\gamma}$ from Section 3.1:
$t_{s}$ (translations in $\operatorname{Re}(\zeta_{2})$), $r_{\theta}$
(rotations in $\zeta_{1}$), and $\delta_{a}$ (non-isotropic dilations). These
maps transform $\ell(z,\zeta)$ in the following way:

(A.4a) $\displaystyle\ell(t_{s}(z),t_{s}(\zeta))$
$\displaystyle=\ell(z,\zeta),\qquad\qquad s\in{\mathbb{R}}$ (A.4b)
$\displaystyle\ell(r_{\theta}(z),r_{\theta}(\zeta))$
$\displaystyle=\ell(z,\zeta),\qquad\qquad\theta\in[0,2\pi)$ (A.4c)
$\displaystyle\ell(\delta_{a}(z),\delta_{a}(\zeta))$
$\displaystyle=a^{-2\gamma}\ell(z,\zeta),\qquad a>0.$ The measure $\mu_{r}$ is
easily seen to be invariant under such translations and rotations. Further
computation reveals that (A.4d) $\displaystyle\delta_{a}^{*}\mu_{r}(\zeta)$
$\displaystyle=a^{\gamma+r+1}\mu_{r}(\zeta).$

###  A.2. $L^{2}$-norms of the kernel function

For $r\in{\mathbb{R}}$, define $\Theta_{r}:\Omega_{\gamma}\to[0,\infty]$ by

(A.5) $\Theta_{r}(z):=\int_{M_{\gamma}}|\ell(z,\zeta)|^{2}\,\mu_{r}(\zeta).$

Applying $t_{s}$ and $r_{\theta}$ for appropriate values of $s$ and $\theta$,
we see from (A.4a) and (A.4b) that for $z\in\Omega_{\gamma}$,

(A.6)
$\Theta_{r}(z)=\Theta_{r}(z_{1},z_{2})=\Theta_{r}(|z_{1}|,i\,\text{Im}(z_{2})).$

Also observe that for $a>0$, (A.4c) and (A.4d) imply

$\displaystyle\Theta_{r}(\delta_{a}(z))=\int_{M_{\gamma}}|\ell(\delta_{a}(z),\zeta)|^{2}\,\mu_{r}(\zeta)$
$\displaystyle=\int_{M_{\gamma}}|\ell(\delta_{a}(z),\delta_{a}(\zeta))|^{2}\,\delta_{a}^{*}\mu_{r}(\zeta)$
(A.7) $\displaystyle=a^{r+1-3\gamma}\,\Theta_{r}(z).$

Now let $\epsilon>0$ be the unique number with $z\in M_{\gamma}^{\epsilon}$.
If $z_{1}\neq 0$, set $a=|z_{1}|^{-1}$ and combine (A.6) and (A.7) to see

$\displaystyle\Theta_{r}(z)=\Theta_{r}(|z_{1}|,i\,\text{Im}(z_{2}))$
$\displaystyle=a^{3\gamma-r-1}\Theta_{r}(\delta_{a}(|z_{1}|,i\,\text{Im}(z_{2})))$
$\displaystyle=|z_{1}|^{r+1-3\gamma}\Theta_{r}\Big{(}1,i\,\tfrac{\text{Im}(z_{2})}{|z_{1}|^{\gamma}}\Big{)}$
(A.8)
$\displaystyle=|z_{1}|^{r+1-3\gamma}\Theta_{r}\Big{(}1,i\left(1+\tfrac{\epsilon}{|z_{1}|^{\gamma}}\right)\Big{)}.$

Otherwise if $z_{1}=0$,

(A.9)
$\Theta_{r}(z)=\Theta_{r}(0,i\,\text{Im}{z_{2}})=\Theta_{r}(0,i\epsilon).$

Now endow $\Omega_{\gamma}$ with a coordinate system based on the
$(\alpha,\theta,s)$-coordinatization of $M_{\gamma}$ that was introduced in
Section 3.2. For $z\in M_{\gamma}^{\epsilon}$, write

(A.10) $\displaystyle z$ $\displaystyle=(\alpha
e^{i\theta},s+i(\alpha^{\gamma}+\epsilon)).$

Points in $\Omega_{\gamma}$ may now be described in
$(\alpha,\theta,s,\epsilon)$-coordinates, where $\alpha\geq 0$,
$\theta\in[0,2\pi)$, $s\in{\mathbb{R}}$, $\epsilon>0$. In these coordinates,
denote the quantity appearing in the innermost set of parentheses of (A.8) as
follows:

(A.11)
$1+\frac{\epsilon}{|z_{1}|^{\gamma}}=1+\frac{\epsilon}{\alpha^{\gamma}}:=q(\alpha,\epsilon).$

The goal is now to understand the function $z\mapsto\Theta_{r}(z)$ from the
perspective of

(A.12)
$(\alpha,\theta,s,\epsilon)\mapsto\alpha^{r+1-3\gamma}\Theta_{r}(1,iq(\alpha,\epsilon)).$

######  Proposition A.13.

Let $r\in{\mathcal{J}}^{\prime}_{0}:=(-1,3\gamma-1)$ and $q>1$. Then

$\Theta_{r}\left(1,iq\right)=q^{\frac{1+r}{\gamma}-3}F_{r}(q),$

where $F_{r}$ is positive, strictly decreasing and real-analytic on
$(1,\infty)$. Furthermore,

  * $(a)$

$\lim_{q\to 1^{+}}F_{r}(q)=\infty.$

  * $(b)$

$\lim_{q\to\infty}F_{r}(q)$ is a positive number.

If $r\notin{\mathcal{J}}^{\prime}_{0}$, then
$\Theta_{r}\left(1,iq\right)=\infty$.

###### Proof.

Denote $\zeta\in M_{\gamma}$ by
$\zeta=(\alpha_{\zeta}e^{i\theta_{\zeta}},s_{\zeta}+i\alpha_{\zeta}^{\gamma})$,
which yields

$\ell((1,iq),\zeta)=\frac{4}{\big{(}(\gamma-1)\alpha_{\zeta}^{\gamma}+q+is_{\zeta}-\gamma\alpha_{\zeta}^{\gamma-1}e^{-i\theta_{\zeta}}\big{)}^{2}}.$

Therefore

$\displaystyle\Theta_{r}(1,iq)$
$\displaystyle=16\int_{M_{\gamma}}\frac{\alpha_{\zeta}^{r}\,ds_{\zeta}\wedge
d\alpha_{\zeta}\wedge
d\theta_{\zeta}}{\big{|}(\gamma-1)\alpha_{\zeta}^{\gamma}+q+is_{\zeta}-\gamma\alpha_{\zeta}^{\gamma-1}e^{-i\theta_{\zeta}}\big{|}^{4}}$
(A.14)
$\displaystyle=16\int_{0}^{\infty}\alpha_{\zeta}^{r}\int_{-\infty}^{\infty}\left\\{\int_{0}^{2\pi}\frac{d\theta_{\zeta}}{|A-Be^{-i\theta_{\zeta}}|^{4}}\right\\}\,ds_{\zeta}\,d\alpha_{\zeta},$

where

(A.15) $A=(\gamma-1)\alpha_{\zeta}^{\gamma}+q+is_{\zeta},\qquad
B=\gamma\alpha_{\zeta}^{\gamma-1}.$

Since $q>1$, Lemma 3.11 shows that $|B|<|A|$ for all choices of
$\alpha_{\zeta},s_{\zeta}$. Focus now on the innermost integral in (A.14):

$\displaystyle\int_{0}^{2\pi}\frac{d\theta_{\zeta}}{|A-Be^{i\theta_{\zeta}}|^{4}}$
$\displaystyle=\int_{0}^{2\pi}\frac{1}{(A-Be^{-i\theta_{\zeta}})^{2}}\cdot\overline{\frac{1}{(A-Be^{-i\theta_{\zeta}})^{2}}}\,d\theta_{\zeta}$
$\displaystyle=\frac{1}{|A|^{4}}\int_{0}^{2\pi}\sum_{j,k=0}^{\infty}(j+1)(k+1)\left(\frac{B}{A}\right)^{j}\overline{\left(\frac{B}{A}\right)^{k}}e^{i(k-j)\theta_{\zeta}}\,d\theta_{\zeta}$
(A.16)
$\displaystyle=\frac{2\pi}{|A|^{4}}\sum_{k=0}^{\infty}(k+1)^{2}\left|\frac{B}{A}\right|^{2k}.$

Plugging (A.16) into (A.14) gives

$\displaystyle\Theta_{r}(1,iq)$
$\displaystyle=32\pi\sum_{k=0}^{\infty}(k+1)^{2}\int_{0}^{\infty}\alpha_{\zeta}^{r}\left|B^{2k}\right|\left\\{\int_{-\infty}^{\infty}\frac{ds_{\zeta}}{|A|^{2k+4}}\right\\}\,d\alpha_{\zeta}$
(A.17)
$\displaystyle=32\pi\sum_{k=0}^{\infty}(k+1)^{2}\gamma^{2k}\int_{0}^{\infty}\alpha_{\zeta}^{2k(\gamma-1)+r}\left\\{\int_{-\infty}^{\infty}\frac{ds_{\zeta}}{|(\gamma-1)\alpha_{\zeta}^{\gamma}+q+is_{\zeta}|^{2k+4}}\right\\}\,d\alpha_{\zeta}.$

We now import the first of two integral formulas which are established in
Section A.4. Proposition A.35 shows that for any $C>0$,

(A.18)
$\int_{-\infty}^{\infty}\frac{ds}{|C+is|^{2k+4}}=\frac{\pi}{4^{k+1}C^{2k+3}}\frac{\Gamma(2k+3)}{\Gamma(k+2)^{2}}.$

Substituting this into (A.17) with $C=(\gamma-1)\alpha_{\zeta}^{\gamma}+q$
shows

(A.19)
$\Theta_{r}(1,iq)=8\pi^{2}\sum_{k=0}^{\infty}\frac{\Gamma(2k+3)}{\Gamma(k+1)^{2}}\left(\frac{\gamma}{2}\right)^{2k}\int_{0}^{\infty}\frac{\alpha_{\zeta}^{2k(\gamma-1)+r}}{((\gamma-1)\alpha_{\zeta}^{\gamma}+q)^{2k+3}}d\alpha_{\zeta}.$

For each fixed $k$, two conditions on $r$ are required for the integral in
(A.19) to converge:

  * (1)

For $\alpha_{\zeta}$ near $0$, we need $2k(\gamma-1)+r>-1$.

  * (2)

As $\alpha_{\zeta}$ tends to $\infty$, we need
$\gamma(2k+3)-2k(\gamma-1)-r>1$.

These conditions suggest the definition of the following interval of $r$
values

(A.20) ${\mathcal{J}}^{\prime}_{k}:=(-2k(\gamma-1)-1,3\gamma-1+2k).$

Note that ${\mathcal{J}}^{\prime}_{0}\subset{\mathcal{J}}^{\prime}_{k}$ for
all $k$, so the convergence of the $k=0$ integral implies the convergence of
all other integrals in the sum. But if the $k=0$ integral diverges, i.e.,
$r\notin{\mathcal{J}}^{\prime}_{0}$, then $\Theta_{r}(1,iq)=\infty$. Now make
use of the second integral formula established in Section A.4. Proposition
A.37 shows that, for $q>0$, $\gamma>1$, $r\in{\mathcal{J}}^{\prime}_{k}$ and
$k$ a non-negative integer,

(A.21)
$\int_{0}^{\infty}\frac{\alpha^{2k(\gamma-1)+r}}{((\gamma-1)\alpha^{\gamma}+q)^{2k+3}}d\alpha=\frac{\Gamma\big{(}3+\frac{2k-1-r}{\gamma}\big{)}\Gamma\big{(}2k+\frac{1+r-2k}{\gamma}\big{)}}{\gamma(\gamma-1)^{\frac{1+r}{\gamma}}q^{3-\frac{1+r}{\gamma}}\Gamma\big{(}2k+3\big{)}}\left(q(\gamma-1)^{\gamma-1}\right)^{-\frac{2k}{\gamma}}.$

Plugging this into (A.19) yields

(A.22)
$\Theta_{r}(1,iq)=\frac{8\pi^{2}}{\gamma(\gamma-1)^{\frac{1+r}{\gamma}}}q^{\frac{1+r}{\gamma}-3}\sum_{k=0}^{\infty}\tfrac{\Gamma\left(3+\frac{2k-1-r}{\gamma}\right)\Gamma\left(2k+\frac{1+r-2k}{\gamma}\right)}{\Gamma\big{(}k+1\big{)}^{2}}\left(\frac{\gamma^{2}}{4(\gamma-1)^{2-\frac{2}{\gamma}}q^{\frac{2}{\gamma}}}\right)^{k}.$

We study the convergence of this sum by considering (for
$r\in{\mathcal{J}}^{\prime}_{k}$) the related series

(A.23) $\sum_{k=0}^{\infty}a_{k}x^{k},\qquad\textrm{where}\qquad
a_{k}=\frac{\Gamma\big{(}3+\frac{2k-1-r}{\gamma}\big{)}\Gamma\big{(}2k+\frac{1+r-2k}{\gamma}\big{)}}{\Gamma\big{(}k+1\big{)}^{2}}.$

Sterling’s formula says

$\displaystyle\Gamma({k+1})^{2}$ $\displaystyle\sim 2\pi e^{-2k}k^{2k+1},$
$\displaystyle\Gamma\big{(}3+\tfrac{2k-1-r}{\gamma}\big{)}$
$\displaystyle\sim\sqrt{2\pi}\cdot
e^{\frac{1+r-2k}{\gamma}-2}\left(2+\tfrac{2k-1-r}{\gamma}\right)^{\frac{5}{2}+\frac{2k-1-r}{\gamma}},$
$\displaystyle\Gamma\big{(}2k+\tfrac{1+r-2k}{\gamma}\big{)}$
$\displaystyle\sim\sqrt{2\pi}\cdot
e^{1-2k-\frac{1+r-2k}{\gamma}}\left(2k-1+\tfrac{1+r-2k}{\gamma}\right)^{2k-\frac{1}{2}+\frac{1+r-2k}{\gamma}}.$

Combining these shows that

$\displaystyle a_{k}$
$\displaystyle\sim\frac{1}{ek^{2k+1}}\cdot\left(2+\tfrac{2k-1-r}{\gamma}\right)^{\frac{5}{2}+\frac{2k-1-r}{\gamma}}\left(2k-1+\tfrac{1+r-2k}{\gamma}\right)^{2k-\frac{1}{2}+\frac{1+r-2k}{\gamma}}$
$\displaystyle=\tfrac{k}{e}\cdot\left(\tfrac{2}{\gamma}+\tfrac{2\gamma-1-r}{\gamma
k}\right)^{\frac{5}{2}+\frac{2k-1-r}{\gamma}}\left(2-\tfrac{2}{\gamma}+\tfrac{1+r-\gamma}{\gamma
k}\right)^{2k-\frac{1}{2}+\frac{1+r-2k}{\gamma}}$
$\displaystyle\sim\tfrac{k}{e}\cdot\left(\tfrac{2}{\gamma}\right)^{\frac{5}{2}-\frac{1+r}{\gamma}}\left(\tfrac{2}{\gamma}+\tfrac{2\gamma-1-r}{\gamma
k}\right)^{\frac{2k}{\gamma}}\left(2-\tfrac{2}{\gamma}\right)^{\frac{1+r}{\gamma}-\frac{1}{2}}\left(2-\tfrac{2}{\gamma}+\tfrac{1+r-\gamma}{\gamma
k}\right)^{2k-\frac{2k}{\gamma}}$
$\displaystyle=\tfrac{k}{e}\cdot\left(\tfrac{2}{\gamma}\right)^{2}(\gamma-1)^{\frac{1+r}{\gamma}-\frac{1}{2}}\left(\tfrac{2}{\gamma}\right)^{\frac{2k}{\gamma}}\left(1+\left(\tfrac{2\gamma-1-r}{\gamma}\right)\tfrac{\gamma}{2k}\right)^{\frac{2k}{\gamma}}$
(A.24)
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\cdot\left(2-\tfrac{2}{\gamma}\right)^{2k-\frac{2k}{\gamma}}\left(1+\left(\tfrac{1+r-\gamma}{\gamma}\right)\left(2k-\tfrac{2k}{\gamma}\right)^{-1}\right)^{2k-\frac{2k}{\gamma}}.$

Since $\lim_{t\to\infty}\left(1+\tfrac{c}{t}\right)^{t}=e^{c}$, we see from
(A.24) that

$\displaystyle a_{k}$
$\displaystyle\sim\tfrac{k}{e}\cdot\left(\tfrac{2}{\gamma}\right)^{2}(\gamma-1)^{\frac{1+r}{\gamma}-\frac{1}{2}}\left(\tfrac{2}{\gamma}\right)^{\frac{2k}{\gamma}}e^{\frac{2\gamma-1-r}{\gamma}}\left(2-\tfrac{2}{\gamma}\right)^{2k-\frac{2k}{\gamma}}e^{\frac{1+r-\gamma}{\gamma}}$
(A.25)
$\displaystyle=4\gamma^{-2}(\gamma-1)^{\frac{1+r}{\gamma}-\frac{1}{2}}k\left(4\gamma^{-2}(\gamma-1)^{2-\frac{2}{\gamma}}\right)^{k}.$

It is easy to see that
$\lim_{k\to\infty}\sqrt[k]{a_{k}}=4\gamma^{-2}(\gamma-1)^{2-\frac{2}{\gamma}}$,
so the root test guarantees the convergence of $\sum_{k=0}^{\infty}a_{k}x^{k}$
on the interval

(A.26) $0\leq x<\frac{\gamma^{2}}{4(\gamma-1)^{2-\frac{2}{\gamma}}}.$

Also from (A.25), it is immediate that $\sum_{k=0}^{\infty}a_{k}x^{k}$
diverges at the right endpoint of (A.26), since the sum $\sum_{k=0}^{\infty}k$
diverges.

These observations about $\sum a_{k}x^{k}$ yield immediate analogues for
$\Theta_{r}(1,iq)$. We now see that, for any $r\in{\mathcal{J}}^{\prime}_{0}$,
formula (A.22) converges for $q>1$. Indeed,

$\Theta_{r}(1,iq)=q^{\frac{1+r}{\gamma}-3}F_{r}(q),$

where

(A.27)
$F_{r}(q)=\frac{8\pi^{2}}{\gamma(\gamma-1)^{\frac{1+r}{\gamma}}}\sum_{k=0}^{\infty}\frac{\Gamma\big{(}3+\frac{2k-1-r}{\gamma}\big{)}\Gamma\big{(}2k+\frac{1+r-2k}{\gamma}\big{)}}{\Gamma\big{(}k+1\big{)}^{2}}\left(\frac{\gamma^{2}}{4(\gamma-1)^{2-\frac{2}{\gamma}}q^{\frac{2}{\gamma}}}\right)^{k}.$

It is now clear that $F_{r}$ is positive, strictly decreasing and real-
analytic on $(1,\infty)$. Furthermore, $\lim_{q\to 1^{+}}F_{r}(q)=\infty$. It
is also clear that

(A.28)
$\lim_{q\to\infty}F_{r}(q)=\frac{8\pi^{2}\Gamma\big{(}3-\frac{1+r}{\gamma}\big{)}\Gamma\big{(}\frac{1+r}{\gamma}\big{)}}{\gamma(\gamma-1)^{\frac{1+r}{\gamma}}}.$

This concludes the proof of Proposition A.13. ∎

Equation (A.8) in conjunction with Proposition A.13 gives a useful description
of $\Theta_{r}(z)$ when $z_{1}\neq 0$. It still remains to consider the
$z_{1}=0$ case, which by (A.9) amounts to understanding
$\Theta_{r}(0,i\epsilon)$. Observe that

$\displaystyle\ell((0,i\epsilon),(\zeta_{1},\zeta_{2}))$
$\displaystyle=\frac{4}{(\gamma|\zeta_{1}|^{\gamma}+\epsilon+is_{\zeta})^{2}}=\frac{4}{((\gamma-1)\alpha_{\zeta}^{\gamma}+\epsilon+is_{\zeta})^{2}},$

and therefore

$\displaystyle\Theta_{r}(0,i\epsilon)$
$\displaystyle=16\int_{M_{\gamma}}\frac{1}{|(\gamma-1)\alpha_{\zeta}^{\gamma}+\epsilon+is_{\zeta}|^{4}}\,\mu_{r}(\zeta)$
$\displaystyle=32\pi\int_{0}^{\infty}\alpha_{\zeta}^{r}\int_{-\infty}^{\infty}\frac{1}{|(\gamma-1)\alpha_{\zeta}^{\gamma}+\epsilon+is_{\zeta}|^{4}}ds_{\zeta}d\alpha_{\zeta}.$

The inner integral is calculated by a special case of (A.18), while the outer
integral is a special case of (A.21). We conclude that

(A.29)
$\Theta_{r}(0,i\epsilon)=\frac{8\pi^{2}\Gamma\big{(}3-\frac{1+r}{\gamma}\big{)}\Gamma\big{(}\frac{1+r}{\gamma}\big{)}}{\gamma(\gamma-1)^{\frac{1+r}{\gamma}}}\,\epsilon^{\frac{1+r}{\gamma}-3}.$

These two results can be incorporated into a single statement. Indeed,
combining (A.12) with Proposition A.13 shows that if $|z_{1}|=\alpha\neq 0$,
$r\in{\mathcal{J}}^{\prime}_{0}$,

$\displaystyle\Theta_{r}(z)=\Theta_{r}(\alpha,\theta,s,\epsilon)=\alpha^{r+1-3\gamma}\Theta_{r}(1,iq(\alpha,\epsilon))$
$\displaystyle=\alpha^{r+1-3\gamma}q(\alpha,\epsilon)^{\frac{r+1}{\gamma}-3}F_{r}(q(\alpha,\epsilon))$
(A.30)
$\displaystyle=(\alpha^{\gamma}+\epsilon)^{\frac{r+1}{\gamma}-3}F_{r}(q(\alpha,\epsilon)).$

Since $z\in M_{\gamma}^{\epsilon}$ for some $\epsilon>0$,
$q(0,\epsilon)=\infty$. Equation (A.28) lets us define
$F_{r}(\infty):=\lim_{q\to\infty}F_{r}(q)$. Comparing (A.29) and (A.30) side
by side shows that the latter now encompasses the former.

######  Remark A.31.

If $\gamma=2$ and $r=1$ (i.e. the Heisenberg group equipped with Euclidean
measure on its parameter space ${\mathbb{R}}^{3}$), the expression for
$\Theta_{r}(z)$ in (A.30) is independent of $\alpha$. This is an instance of a
phenomenon occurring on the ${\mathcal{S}}_{\beta}$ hypersurfaces defined in
(1.12). Refer to the Appendix in [6] for more information. $\lozenge$

###  A.3. Proof of Theorem A.1

Let $\mathscr{B}$ denote the set of all subsets $K\subset\Omega_{\gamma}$
satisfying the following two properties:

  1. (1)

$\alpha_{K}:=\sup\\{\alpha:(\alpha,\theta,s,\epsilon)\in K\\}<\infty$.

  2. (2)

$\epsilon_{K}:=\inf\\{\epsilon:(\alpha,\theta,s,\epsilon)\in K\\}>0$.

Given a fixed $K\in{\mathscr{B}}$, observe that the infimum of
$q(\alpha,\epsilon)$ when restricted to $K$ remains strictly greater than $1$.
Indeed, if we define

(A.32) $q_{K}:=\inf\\{q(\alpha,\epsilon):(\alpha,\theta,s,\epsilon)\in K\\},$

it is clear that $q_{K}\geq 1+\tfrac{\epsilon_{K}}{\alpha_{K}^{\gamma}}>1$.
${\mathscr{B}}$ contains, for instance, all compact subsets of
$\Omega_{\gamma}$; it also contains unbounded subsets of $\Omega_{\gamma}$, as
it places no restriction on $s$.

Fix some $r\in{\mathcal{J}}_{0}=(-1-\gamma,2\gamma-1)$. This determines a
unique $r^{\prime}$ value with $\frac{r+r^{\prime}}{2}=\gamma-1$, and it is
clear that $r^{\prime}\in{\mathcal{J}}^{\prime}_{0}=(-1,3\gamma-1)$. Now
observe that if $K\in{\mathscr{B}}$, then $\sup_{z\in
K}\Theta_{r^{\prime}}(z)$ is bounded from above. Indeed, for $z\in K$, (A.30)
shows

(A.33)
$\Theta_{r^{\prime}}(z)=(\alpha^{\gamma}+\epsilon)^{\frac{r^{\prime}+1}{\gamma}-3}F_{r^{\prime}}(q(\alpha,\epsilon))\leq\epsilon_{K}^{\frac{r^{\prime}+1}{\gamma}-3}F_{r^{\prime}}(q_{K}).$

Proposition A.13 guarantees that this is finite since $q_{K}>1$.

If $f\in L^{2}(M_{\gamma},\mu_{r})$, $r\in{\mathcal{J}}_{0}$, we will show
that $\bm{L}f\in{\mathcal{O}}(\Omega_{\gamma})$ by proving it is holomorphic
on each $K\in{\mathscr{B}}$. First assume $f\in L^{2}(M_{\gamma},\mu_{r})$ is
compactly supported. Then $\bm{L}f$ is seen to be holomorphic by
differentiating under the integral sign in (A.3). For a general $f\in
L^{2}(M_{\gamma},\mu_{r})$, choose a sequence of compactly supported $f_{j}$
tending to $f$ in $L^{2}(M_{\gamma},\mu_{r})$.

$\displaystyle\sup_{z\in K}\left|\bm{L}f_{j}(z)-\bm{L}f(z)\right|$
$\displaystyle=\frac{\gamma^{2}}{16\pi^{2}}\sup_{z\in
K}\left|\int_{M_{\gamma}}\ell(z,\zeta)(f_{j}(\zeta)-f(\zeta))\sigma(\zeta)\right|$
$\displaystyle\leq\frac{\gamma^{2}}{16\pi^{2}}\sup_{z\in
K}\left(\int_{M_{\gamma}}|\ell(z,\zeta)|^{2}\mu_{r^{\prime}}(\zeta)\right)\left(\int_{M_{\gamma}}|f_{j}(\zeta)-f(\zeta)|^{2}\mu_{r}(\zeta)\right)$
$\displaystyle=C_{r^{\prime},K}\left\|f_{j}-f\right\|_{\mu_{r}}^{2},$

where the constant only depends on $r^{\prime}$ and $K$. This shows that
$\bm{L}f_{j}\to\bm{L}f$ uniformly on $K$, implying
$\bm{L}f\in{\mathcal{O}}(K)$, being a uniform limit of holomorphic functions.
Since $K\in{\mathscr{B}}$ was arbitrary, we conclude that $\bm{L}f$ is
holomorphic on all of $\Omega_{\gamma}$. ∎

######  Remark A.34.

Calculations seen throughout this appendix can be easily adapted to the sub-
Leray operator $\bm{L}_{k}$. Suppose that
$r^{\prime}\in{\mathcal{J}}^{\prime}_{k}=(-2k(\gamma-1)-1,3\gamma-1+2k)$, the
interval defined in (A.20). If $\frac{r+r^{\prime}}{2}=\gamma-1$, it holds
that $r\in(-\gamma-1-2k,2\gamma-1+2k(\gamma-1)):={\mathcal{J}}_{k}$. It can be
verified that if $r\in{\mathcal{J}}_{k}$ and $f\in L^{2}(M_{\gamma},\mu_{r})$,
then $\bm{L}_{k}f\in{\mathcal{O}}(\Omega_{\gamma})$. Recall now that the
interval of $r$ values for which $\bm{L}_{k}$ is bounded from
$L^{2}(M_{\gamma},\mu_{r})\to L^{2}(M_{\gamma},\mu_{r})$ was given in (5.12)
by ${\mathcal{I}}_{k}=(-2k-1,(2k+2)(\gamma-1)+1)$. It is immediate that
${\mathcal{I}}_{k}\subset{\mathcal{J}}_{k}$. $\lozenge$

###  A.4. Two residue integral computations

######  Proposition A.35.

For $C>0$ and $k$ a non-negative integer,

$\int_{-\infty}^{\infty}\frac{ds}{|C+is|^{2k+4}}=\frac{\pi}{4^{k+1}C^{2k+3}}\cdot\frac{\Gamma(2k+3)}{\Gamma(k+2)^{2}}.$

###### Proof.

Write

$\int_{-\infty}^{\infty}\frac{ds}{|C+is|^{2k+4}}=\int_{-\infty}^{\infty}\frac{ds}{(s^{2}+C^{2})^{k+2}}$

Let $\mathcal{P}_{R}$ denote a counter-clockwise oriented, closed half-circle
contour with radius $R$ lying in the upper-half plane with base lying on the
interval $[-R,R]$. If we set $g(z)={(z^{2}+C^{2})^{-k-2}}$, the residue
theorem says (provided $R>C$),

(A.36) $2\pi i\,\textrm{Res}(g,iC)=\int_{\mathcal{P}_{R}}g(z)\,dz.$

We calculate

$\displaystyle\textrm{Res}(g,iC)$
$\displaystyle=\frac{1}{\Gamma(k+2)}\frac{d^{k+1}}{dz^{k+1}}\left[\frac{1}{(z+iC)^{k+2}}\right]\Bigg{|}_{z=iC}$
$\displaystyle=\frac{1}{i2^{2k+3}C^{2k+3}}\cdot\frac{\Gamma(2k+3)}{\Gamma(k+2)^{2}}.$

Now send $R\to\infty$ and note that the circular portion of the contour
integral in (A.36) tends to $0$. This gives the result. ∎

######  Proposition A.37.

Let $q>0$, $\gamma>1$,
$r\in{\mathcal{J}}^{\prime}_{k}=(-2k(\gamma-1)-1,3\gamma-1+2k)$ and $k$ a non-
negative integer. Then

$\int_{0}^{\infty}\frac{\alpha^{2k(\gamma-1)+r}}{((\gamma-1)\alpha^{\gamma}+q)^{2k+3}}d\alpha=\frac{\Gamma\left(3+\frac{2k-1-r}{\gamma}\right)\Gamma\left(2k+\frac{1+r-2k}{\gamma}\right)}{\gamma(\gamma-1)^{\frac{1+r}{\gamma}}q^{3-\frac{1+r}{\gamma}}\Gamma\big{(}2k+3\big{)}}\left(q(\gamma-1)^{\gamma-1}\right)^{-\frac{2k}{\gamma}}.$

###### Proof.

Making the change of variable $x=\alpha^{\gamma}$, see that

(A.38)
$\displaystyle\int_{0}^{\infty}\frac{\alpha^{2k(\gamma-1)+r}}{((\gamma-1)\alpha^{\gamma}+q)^{2k+3}}\,d\alpha$
$\displaystyle=\frac{1}{\gamma(\gamma-1)^{2k+3}}\int_{0}^{\infty}\frac{x^{2k-1+\frac{r+1-2k}{\gamma}}}{(x+E)^{2k+3}}\,dx,$

with $E=\frac{q}{\gamma-1}$. Now define the function

(A.39) $h(z)=\frac{z^{2k-1+\frac{r+1-2k}{\gamma}}}{(z+E)^{2k+3}},$

which is multi-valued whenever the numerator exponent is a non-integer.
Restrict attention to a single-valued meromorphic branch of $h$ defined on
${\mathbb{C}}\backslash[0,\infty)$. Note that $h$ has a pole of order $2k+3$
at $z=-E$.

For $0<\delta<R$, define the closed, positively oriented contour
$\mathcal{P}_{\delta,R}$ by traversing the following paths in sequence:

  * $I$.

A line segment on the positive $x$-axis moving right from $\delta$ to $R$.

  * $II$.

A circle of radius $R$ centered at the origin, starting on the positive
$x$-axis and traveling counterclockwise.

  * $III$.

A line segment on the positive $x$-axis moving left from $R$ to $\delta$.

  * $IV$.

A circle of radius $\delta$ centered at the origin, starting on the positive
$x$-axis and traveling clockwise.

As long as $\delta<E<R$, the residue theorem says

$\displaystyle 2\pi i\,\textrm{Res}(h,-E)$
$\displaystyle=\int_{\mathcal{P}_{\delta,R}}h(z)\,dz$ (A.40)
$\displaystyle=\int_{I}h(z)\,dz+\int_{II}h(z)\,dz+\int_{III}h(z)\,dz+\int_{IV}h(z)\,dz.$

Keeping in mind the multi-valued nature of $h$, combine the $I$ and $III$
integrals:

$\displaystyle\int_{I}h(z)\,dz+\int_{III}h(z)\,dz$
$\displaystyle=\int_{\delta}^{R}h(x)\,dx+\int^{\delta}_{R}h(xe^{2\pi i})\,dx$
$\displaystyle=\int_{\delta}^{R}\frac{x^{2k-1+\frac{r+1-2k}{\gamma}}}{(x+E)^{2k+3}}\,dx+\int^{\delta}_{R}\frac{x^{2k-1+\frac{r+1-2k}{\gamma}}e^{2\pi
i(2k-1+\frac{r+1-2k}{\gamma})}}{(xe^{2\pi i}+E)^{2k+3}}\,dx$ (A.41)
$\displaystyle=\left(1-e^{\frac{2\pi
i(r+1-2k)}{\gamma}}\right)\int_{\delta}^{R}\frac{x^{2k-1+\frac{r+1-2k}{\gamma}}}{(x+E)^{2k+3}}\,dx.$

Standard estimates show that since $r\in{\mathcal{J}}^{\prime}_{k}$,

(A.42) $\lim_{R\to\infty}\int_{II}h(z)\,dz=0=\lim_{\delta\to
0}\int_{IV}h(z)\,dz.$

Combining (A.42) with (A.40) and (A.41) shows

(A.43) $2\pi i\,\textrm{Res}(h,-E)=\left(1-e^{\frac{2\pi
i(r+1-2k)}{\gamma}}\right)\int_{0}^{\infty}\frac{x^{2k-1+\frac{r+1-2k}{\gamma}}}{(x+E)^{2k+3}}\,dx.$

Now calculate this residue.

$\displaystyle\textrm{Res}(h,-E)$
$\displaystyle=\frac{1}{\Gamma(2k+3)}\frac{d^{2k+2}}{dz^{2k+2}}\left(z^{2k-1+\frac{r+1-2k}{\gamma}}\right)\Bigg{|}_{z=-E}$
(A.44)
$\displaystyle=\frac{-\Gamma\big{(}{2k+\frac{r+1-2k}{\gamma}\big{)}}}{\Gamma\big{(}2k+3\big{)}\Gamma\big{(}\frac{r+1-2k}{\gamma}-2\big{)}}E^{\frac{r+1-2k}{\gamma}-3}e^{\frac{\pi
i(r+1-2k)}{\gamma}}.$

Combining (A.43) and (A.44) shows that

$\displaystyle\int_{0}^{\infty}\frac{x^{2k(1-\frac{1}{\gamma})}}{(x+E)^{2k+3}}\,dx$
$\displaystyle=\frac{-2\pi i}{1-e^{\frac{2\pi
i(r+1-2k)}{\gamma}}}\cdot\frac{\Gamma\big{(}{2k+\frac{r+1-2k}{\gamma}\big{)}}}{\Gamma\big{(}2k+3\big{)}\Gamma\big{(}\frac{r+1-2k}{\gamma}-2\big{)}}E^{\frac{r+1-2k}{\gamma}-3}e^{\frac{\pi
i(r+1-2k)}{\gamma}}$
$\displaystyle=\pi\csc\left(\pi\big{(}\tfrac{r+1-2k}{\gamma}\big{)}\right)\frac{\Gamma\big{(}{2k+\frac{r+1-2k}{\gamma}\big{)}}}{\Gamma\big{(}2k+3\big{)}\Gamma\big{(}\frac{r+1-2k}{\gamma}-2\big{)}}E^{\frac{r+1-2k}{\gamma}-3}$
$\displaystyle=\frac{\Gamma\big{(}\frac{r+1-2k}{\gamma}\big{)}\Gamma\big{(}1-\frac{r+1-2k}{\gamma}\big{)}\Gamma\big{(}2k+\frac{r+1-2k}{\gamma}\big{)}}{\Gamma\big{(}2k+3\big{)}\Gamma\big{(}\frac{r+1-2k}{\gamma}-2\big{)}}E^{\frac{r+1-2k}{\gamma}-3}$
$\displaystyle=\big{(}\tfrac{r+1-2k}{\gamma}-1\big{)}\big{(}\tfrac{r+1-2k}{\gamma}-2\big{)}\frac{\Gamma\big{(}1-\frac{r+1-2k}{\gamma}\big{)}\Gamma\big{(}2k+\frac{r+1-2k}{\gamma}\big{)}}{\Gamma\big{(}2k+3\big{)}}E^{\frac{r+1-2k}{\gamma}-3}$
(A.45)
$\displaystyle=\frac{\Gamma\big{(}3+\frac{2k-r-1}{\gamma}\big{)}\Gamma\big{(}2k+\frac{r+1-2k}{\gamma}\big{)}}{\Gamma\big{(}2k+3\big{)}}E^{\frac{r+1-2k}{\gamma}-3}.$

Combining (A.38) and (A.45) gives the result. ∎

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