# The first partial derivatives of generalized harmonic functions

Adel Khalfallah  Department of Mathematics, King Fahd University of Petroleum
and Minerals, Dhahran 31261, Saudi Arabia<EMAIL_ADDRESS>and  Mohamed
Mhamdi  Ecole supérieure des Sciences et de la Technologie de Hammam Sousse
(ESSTHS)
Université de Sousse<EMAIL_ADDRESS>
###### Abstract.

Suppose ${\alpha},{\beta}\in{\mathbb{R}}\backslash{\mathbb{Z}}^{-}$ such that
${\alpha}+{\beta}>-1$ and $1\leq p\leq\infty$. Let $u=P_{{\alpha},{\beta}}[f]$
be an $({\alpha},{\beta})$-harmonic mapping on ${\mathbb{D}}$, the unit disc
of ${\mathbb{C}}$, with the boundary $f$ being absolutely continuous and
$\dot{f}\in L^{p}(0,2\pi)$, where
$\dot{f}(e^{i\theta}):=\frac{d}{d\theta}f(e^{i\theta})$. In this paper, we
investigate the membership of the partial derivatives $\partial_{z}u$ and
$\partial_{\overline{z}}u$ in the space $H_{G}^{p}(\mathbb{D})$, the
generalized Hardy space. We prove, if ${\alpha}+{\beta}>0$, then both
$\partial_{z}u$ and $\partial_{\overline{z}}u$ are in $H_{G}^{p}(\mathbb{D})$.
For ${\alpha}+{\beta}<0$, we show if $\partial_{z}u$ or
$\partial_{\overline{z}}u\in H_{G}^{1}(\mathbb{D})$ then $u=0$ or $u$ is a
polyharmonic function.

###### Key words and phrases:

Hardy spaces, $({\alpha},{\beta})$-harmonic functions, Poisson integral

###### 1991 Mathematics Subject Classification:

Primary: 31A30; Secondary: 31A05, 35J25

††footnotetext:  Date: 30-8-2024.

##  1\. Introduction

Let ${\mathbb{D}}$ be the open unit disc in the complex plane ${\mathbb{C}}$,
let ${\mathbb{T}}=\partial{\mathbb{D}}$ be the unit circle, and denote by

$\partial_{z}=\frac{1}{2}(\partial_{x}-i\partial_{y})\quad\quad\text{and}\quad\quad\partial_{\overline{z}}=\frac{1}{2}(\partial_{x}+i\partial_{y}).$

###  1.1. $({\alpha},{\beta})$-harmonic functions

Consider the family of second order partial differential operators on
${\mathbb{D}}$ of the form

$\Delta_{{\alpha},{\beta}}:=(1-|z|^{2})\left((1-|z|^{2})\partial_{z}\partial_{\overline{z}}+{\alpha}z\partial_{z}+{\beta}\,\overline{z}\partial_{\overline{z}}-{\alpha}{\beta}\right),$

where ${\alpha},{\beta}$ are real parameters. These operators have been
introduced and studied in higher dimensions by Geller [10] and Ahern et al.
[1, 2] and recently investigated in the planar case in [15].

Of particular interest are the solutions of the associated homogeneous
equation in ${\mathbb{D}}$:

$\Delta_{{\alpha},{\beta}}\,u=0.$ (1.1)

We say that a function $u$ is $({\alpha},{\beta})$-harmonic if $u$ is twice
continuously differentiable in ${\mathbb{D}}$ and
$\Delta_{{\alpha},{\beta}}u=0$. We remark if $u$ is
$({\alpha},{\beta})$-harmonic, then $\overline{u}$ is
$({\beta},{\alpha})$-harmonic.
$(0,{\alpha})$-harmonic functions are called $\alpha$-harmonic functions.

An interesting example of an $({\alpha},{\beta})$-harmonic function is
$K_{{\alpha},{\beta}}$ defined by

$K_{{\alpha},{\beta}}(z)=\frac{(1-|z|^{2})^{{\alpha}+{\beta}+1}}{(1-z)^{{\alpha}+1}(1-\overline{z})^{{\beta}+1}}.$

For ${\alpha},{\beta}\in{\mathbb{R}}\backslash{\mathbb{Z}}^{-}$ such that
${\alpha}+{\beta}>-1$, the $({\alpha},{\beta})-$harmonic Poisson kernel is
defined by

$P_{{\alpha},{\beta}}(z)=c_{{\alpha},{\beta}}\,\frac{(1-|z|^{2})^{{\alpha}+{\beta}+1}}{(1-z)^{{\alpha}+1}(1-\overline{z})^{{\beta}+1}},\quad
c_{{\alpha},{\beta}}=\frac{\Gamma({\alpha}+1)\Gamma({\beta}+1)}{\Gamma({\alpha}+{\beta}+1)},$
(1.2)

where $\Gamma$ is the Gamma function.

The $({\alpha},{\beta})-$harmonic Poisson integral of $f\in
L^{1}({\mathbb{T}})$ is defined by

$P_{{\alpha},{\beta}}[f](z):=\frac{1}{2\pi}\int_{0}^{2\pi}P_{{\alpha},{\beta}}(ze^{-i\theta})f(e^{i\theta})d\theta,\quad
z\in{\mathbb{D}}.$

Let

$M_{{\alpha},{\beta}}(z):=P_{{\alpha},{\beta}}[1](z)=\frac{1}{2\pi}\int_{0}^{2\pi}P_{{\alpha},{\beta}}(ze^{-i\theta})\,d\theta.$

Then $M_{{\alpha},{\beta}}$ is a radial function and

$M_{{\alpha},{\beta}}(z)=c_{{\alpha},{\beta}}F(-{\alpha},-{\beta};1;|z|^{2}),$

where $F$ is the Gaussian hypergeometric function. The constant
$c_{{\alpha},{\beta}}$ is chosen such that $\lim_{|z|\to
1}M_{{\alpha},{\beta}}(z)=1$. We should point out that in the case
${\alpha}=0$ or ${\beta}=0$, the function $M_{{\alpha},{\beta}}$ is constant
and $M_{{\alpha},{\beta}}(z)=1$. For more properties and Hölder continuity of
$({\alpha},{\beta})$-harmonic functions, see [1, 5, 14, 15, 16].

Let us recall the Gaussian hypergeometric function defined by

$F(a,b;c;z)=\sum_{n\geq 0}\frac{(a)_{n}(b)_{n}}{(c)_{n}}\frac{z^{n}}{n!},\ \ \
\ z\in{\mathbb{D}},$

for $a,b,c\in{\mathbb{C}}$ such that $c\neq-1,-2,...$ where

$(a)_{n}=\begin{cases}1&\mbox{for}\ \ n=0,\\\ a(a+1)\ldots(a+n-1)&\mbox{for}\
\ n=1,2,3,...\par\end{cases}$

which are called Pochhammer symbols.

We list few properties, see for instance ([3], Chapter 2)

$\displaystyle\lim\limits_{x\rightarrow 1}F(a,b;c;x)$ $\displaystyle=$
$\displaystyle\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)},\ \ \ \ \
\text{if}\ \ \ c-a-b>0,$ (1.3) $\displaystyle F(a,b;c;x)$ $\displaystyle=$
$\displaystyle(1-x)^{c-a-b}F(c-a,c-b;c;x),$ (1.4)
$\displaystyle\frac{d}{dx}F(a,b;c;x)$ $\displaystyle=$
$\displaystyle\frac{ab}{c}F(1+a,1+b;1+c;x),$ (1.5) $\displaystyle
F(a,b;a+b;x)$ $\displaystyle\approx$
$\displaystyle-\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\log(1-x)\quad(x\to
1^{-}).$ (1.6)

The asymptotic relation (1.6) is due to Gauss, and its refined form is due to
Ramanujan [4, p. 71].

Also, we consider the associated Dirichlet boundary value problem of functions
$u$, satisfying the equation (1.1),

$\begin{cases}\Delta_{{\alpha},{\beta}}\,u&=0\quad\text{in}\quad{\mathbb{D}},\\\
u&=f\quad\text{on}\quad{\mathbb{T}}.\end{cases}$ (1.7)

The function $f$ is a continuous on ${\mathbb{T}}$, i.e.,
$f\in\mathcal{C}({\mathbb{T}})$, and the condition (1.7) is
$\displaystyle\lim_{r\to 1}u_{r}=f$ in $\mathcal{C}({\mathbb{T}})$, where
$u_{r}(e^{i\theta}):=u(re^{i\theta}).$

Theorem A ([10]). Let
${\alpha},{\beta}\in{\mathbb{R}}\backslash{\mathbb{Z}}^{-}$ be such that
${\alpha}+{\beta}>-1$. Let $f\in\mathcal{C}({\mathbb{T}})$. Then a function
$u$ in ${\mathbb{D}}$ satisfies (1.7) if and only if it has the form

$u(z)=P_{{\alpha},{\beta}}[f](z)=\frac{1}{2\pi}\int_{0}^{2\pi}P_{{\alpha},{\beta}}(ze^{-i\gamma})f(e^{i\gamma})d\gamma,$

for any $z\in{\mathbb{D}}$.

The homogeneous series expansion of $({\alpha},{\beta})$-harmonic functions is
giving by the following theorem.

Theorem B ([1], [15]). Let ${\alpha},{\beta}\in{\mathbb{R}}$. Then $u$ is an
$({\alpha},{\beta})$-harmonic function if and only if it has the form

$u(z)=\sum_{k\geq 0}c_{k}F(-{\alpha},k-{\beta};k+1;|z|^{2})z^{k}+\sum_{k\geq
1}c_{-k}F(-{\beta},k-{\alpha};k+1;|z|^{2})\overline{z}^{k},$ (1.8)

where $c_{k}\in\mathbb{C}$, for all $k\in\mathbb{Z}$.

Thus for ${\alpha},{\beta}\in{\mathbb{R}}\backslash{\mathbb{Z}}^{-}$ with
${\alpha}+{\beta}>-1$, $f\in\mathcal{C}({\mathbb{T}})$, and
$u=P_{{\alpha},{\beta}}[f]$, the relationship between the coefficients $c_{k}$
in (1.8) and $\hat{f}(k)$, the Fourier coefficients of $f$, is given by

$c_{k}F(-{\alpha},k-{\beta};k+1;1)=\hat{f}(k)\mbox{ for }k\geq 0,$ (1.9)

and

$c_{-k}F(-{\beta},k-{\alpha};k+1;1)=\hat{f}(-k)\mbox{ for }k\geq 1.$ (1.10)

Let $\lambda>-1$ and $r\in[0,1)$, we denote by

$I_{\lambda}(r):=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{(1-r^{2})^{\lambda}}{|1-re^{-i\theta}|^{\lambda+1}}\,d\theta,$

For $\lambda>0$, we have

$\lim_{r\to
1}I_{\lambda}(r)=\frac{\Gamma(\lambda)}{\Gamma^{2}(\frac{\lambda}{2}+\frac{1}{2})}=:c_{\lambda}.$
(1.11)

The following lemma provides some estimates regarding $I_{\lambda}$.

######  Lemma 1.1.

([13, Proposition 1.1] and [17]) Let $r\in[0,1)$.

  1. (1)

If $\lambda>0$, then

$I_{\lambda}(r)\leq c_{\lambda}.$

  2. (2)

If $\ -1<\lambda<0$, then

$I_{\lambda}(r)\leq c_{-\lambda}(1-r^{2})^{\lambda}.$

###  1.2. Hardy type spaces

For $p\in(0,\infty]$, the generalized Hardy space $H_{G}^{p}({\mathbb{D}})$
consists of all measurable functions $u$ from ${\mathbb{D}}$ to ${\mathbb{C}}$
such that $M_{p}(r,u)$ exists for all $r\in(0,1)$, and $\|u\|_{p}<\infty$,
where

$M_{p}(r,u)=\bigg{(}\frac{1}{2\pi}\int_{0}^{2\pi}|u(re^{i\theta})|^{p}d\theta\bigg{)}^{\frac{1}{p}}$

and

$\|u\|_{p}=\begin{cases}\sup\\{M_{p}(r,u):0<r<1\\}&\mbox{if}\ \
p\in(0,\infty)\\\ \sup\\{|u(z)|:z\in{\mathbb{D}}\\}&\mbox{if}\ \
p=\infty.\end{cases}$

Obviously, $H^{p}_{G}({\mathbb{D}})\subset L^{p}({\mathbb{D}})$ for each
$p\in(0,\infty]$. We denote by $h_{{\alpha},{\beta}}^{p}({\mathbb{D}})$ the
corresponding Hardy space for $({\alpha},{\beta})$-harmonic functions. The
classical Hardy space $H^{p}(\mathbb{D})$ (resp. $h^{p}(\mathbb{D})$) is the
set of all elements of $H_{G}^{p}(\mathbb{D})$ which are analytic (resp.
harmonic) on ${\mathbb{D}}$. (cf. [8, 9]).

The Hardy theory of $({\alpha},{\beta})$-harmonic functions is studied in
detail in [1]. The authors proved the following

Theorem C. [1] Let $u$ be an $({\alpha},{\beta})$-harmonic function, and
assume ${\alpha}+{\beta}>-1$. Then :

  1. (i)

$u=P_{{\alpha},{\beta}}[f]$ for some $f\in L^{p}({\mathbb{T}})$,
$1<p\leq\infty$ if and only if $\|u\|_{p}<+\infty$.

  2. (ii)

$u=P_{{\alpha},{\beta}}[\mu]$ for some measure $\mu$ if and only if
$\|u\|_{1}<+\infty$.

###  1.3. Lebesgue spaces

Denote by $L^{p}({\mathbb{T}})$ where $p\in[1,\infty]$ the space of all
measurable functions $f$ of ${\mathbb{T}}$ into ${\mathbb{C}}$ with

$\|f\|_{L^{p}}=\begin{cases}\bigg{(}\displaystyle\frac{1}{2\pi}\int_{0}^{2\pi}|f(e^{i\theta})|^{p}d\theta\bigg{)}^{\frac{1}{p}}&\mbox{if}\
\ p\in[1,\infty)\\\
\text{ess}\sup\\{|f(e^{i\theta})|:\theta\in[0,2\pi)\\}&\mbox{if}\ \
p=\infty.\end{cases}$

It is well-known that if $f$ is absolutely continuous, then it is of bounded
variation. This implies that for almost all $e^{i\theta}\in{\mathbb{T}}$, the
derivative $\dot{f}(e^{i\theta})$ exists and integrable on $(0,2\pi)$, where

$\dot{f}(e^{i\theta}):=\frac{d}{d\theta}f(e^{i\theta}).$

###  1.4. Harmonic conjugates and M. Riesz theorem

If $u$ is a harmonic function on ${\mathbb{D}}$ given by

$u(re^{i\theta})=\sum_{n=-\infty}^{\infty}c_{n}r^{|n|}e^{in\theta},$

then its harmonic conjugate is defined by

$\tilde{u}(re^{i\theta}):=\sum_{n=-\infty}^{\infty}m_{n}c_{n}r^{|n|}e^{in\theta},$

where $m_{n}=-i\,{\rm sign}\,n$; in particular $\tilde{u}(0)=0$. In addition,
we have

$u(z)+i\tilde{u}(z)=-u(0)+2\sum_{n=0}^{\infty}c_{n}z^{n}.$ (1.12)

The expression $\sum_{n\geq 0}c_{n}z^{n}$ is called the Riesz projection of
$u$, and denoted by $P_{+}u$ i.e.,

$P_{+}:h({\mathbb{D}})\to H({\mathbb{D}})$

is defined by

$P_{+}(u)(z):=\sum_{n\geq 0}c_{n}z^{n}.$

Thus Eq. (1.12) can be rewritten as

$u+i\tilde{u}=-u(0)+2P_{+}(u).$

It is natural to consider the following question: if $u$ is harmonic in
${\mathbb{D}}$ and $u\in h^{p}({\mathbb{D}})$, $1\leq p\leq\infty$, does
$P_{+}(u)\in H^{p}({\mathbb{D}})$?
The question has an affirmative answer for all $1<p<\infty$. This is the
content of the famous theorem proved by M. Riesz [18], see also Rudin [19,
Theorem 17.26].

Theorem D. (M. Riesz) For $1<p<\infty$, the Riesz projection $P_{+}$

$P_{+}:h^{p}({\mathbb{D}})\to H^{p}({\mathbb{D}})$

is a bounded operator, i.e., there is a constant $A_{p}$ such that

$\|P_{+}(u)\|_{p}\leq A_{p}\|u\|_{p}$

for any $\ f\in h^{p}({\mathbb{D}})$.  It turns out that the Riesz theorem is
not true for $p=1$ and $p=\infty$.

In this paper, we consider the following :

Question: Under what conditions on the boundary function $f$ ensure that the
partial derivatives of its $({\alpha},{\beta})$-harmonic extension
$u=P_{{\alpha},{\beta}}[f]$, i.e., $\partial_{z}u$ and
$\partial_{\overline{z}}u$, are in the space $H^{p}_{G}({\mathbb{D}})$ (or
$L^{p}({\mathbb{D}})$), where $p\in[1,\infty]$?

In [20], Zhu treated the corresponding problem for harmonic functions and was
improved in [7].

Theorem E ([20] and [7]). Suppose that $u=P[f]$ is a harmonic mapping in
${\mathbb{D}}$ and $\dot{f}\in L^{p}({\mathbb{T}})$, where $f$ is an
absolutely continuous function.

  * •

If $p\in[1,\infty)$, then both $\partial_{z}u$ and $\partial_{\overline{z}}u$
are in $L^{p}({\mathbb{D}})$.

  * •

If $p=\infty$, then there exists a harmonic mapping $u=P[f]$, with $\dot{f}\in
L^{\infty}({\mathbb{T}})$, such that neither $\partial_{z}u$ nor
$\partial_{\overline{z}}u$ is in $L^{\infty}({\mathbb{D}})$.

Furthermore, by requiring the harmonic extension to be quasiregular [20] or
more generally elliptic [7], they proved that the partial derivatives are in
$H^{p}({\mathbb{D}})$, for $p\in[1,\infty]$.

Theorem F ([7]). Suppose that $p\in[1,\infty]$ and $u=P[f]$ is a
$(K,K^{\prime})$-elliptic mapping in ${\mathbb{D}}$ with $\dot{f}\in
L^{p}({\mathbb{T}})$, where $f$ is an absolutely continuous function, $K\geq
1$ and $K^{\prime}\geq 0$. Then both $\partial_{z}u$ and
$\overline{\partial_{\overline{z}}u}$ are in $H^{p}({\mathbb{D}})$.

In [12], we provide a refinement of the two previous theorems, we prove that
for $1<p<\infty$, both $\partial_{z}u$ and
$\overline{\partial_{\overline{z}}u}$ are in $H^{p}(\mathbb{D})$ without any
extra conditions on $u$.

Theorem G. Suppose that $f$ is an absolutely continuous function on
${\mathbb{T}}$ and $u=P[f]$ is a harmonic mapping in $\mathbb{D}$ and
$\dot{f}\in L^{p}(\mathbb{T})$.

  1. (1)

If $p\in(1,\infty)$, then both $\partial_{z}u$ and
$\overline{\partial_{\overline{z}}u}$ are in $H^{p}(\mathbb{D}).$ Moreover,
there exists a constant $A_{p}$ such that

$\max(\|\partial_{z}u\|_{p},\|\partial_{\overline{z}}u\|_{p})\leq
A_{p}\|\dot{f}\|_{L^{p}}.$

  2. (2)

If $p=1$ or $p=\infty$, then both $\partial_{z}u$ and
$\overline{\partial_{\overline{z}}u}$ are in $H^{p}(\mathbb{D})$ if and only
if $H(\dot{f})\in L^{p}({\mathbb{T}})$. Moreover

$2iz\partial_{z}u(z)=P[\dot{f}+iH(\dot{f})](z),$

where $H(\dot{f})$ denotes the Hilbert transform of $\dot{f}$.

In [11], the authors considered the case of $\alpha$-harmonic functions and
they proved the following.

Theorem H. Let $\alpha\in(-1,\infty)$ with $\alpha\not=0$ and let
$u=P_{\alpha}[f]$ and $f$ is absolutely continuous such that $\dot{f}\in
L^{p}({\mathbb{T}})$ with $1\leq p\leq\infty$.

  1. (1)

If $\alpha>0$, then $\partial_{z}u$ and $\partial_{\overline{z}}u$ are in
$H_{G}^{p}({\mathbb{D}})\subset L^{p}({\mathbb{D}})$.

  2. (2)

If $\alpha\in(-1,0)$, then $\partial_{z}u$ and $\partial_{\overline{z}}u$ are
in $L^{p}({\mathbb{D}})$ for $p<-\frac{1}{\alpha}$.

  3. (3)

For $\alpha\in(-1,0)$ and $p\geq-\frac{1}{\alpha}$ there exists $u$ an
$\alpha$-harmonic function such that $\partial_{z}u$ and
$\partial_{\overline{z}}u$ $\not\in L^{p}({\mathbb{D}})$, moreover,
$\partial_{z}u$ and $\partial_{\overline{z}}u$ $\not\in
H_{G}^{1}({\mathbb{D}})$.

  4. (4)

Let $\alpha\in(-1,0)$. Then $\partial_{z}u$ or $\partial_{\overline{z}}u$ is
in $H_{G}^{p}({\mathbb{D}})$, if and only if, $u$ is analytic.

Recall that $P_{\alpha}$ is the Poisson kernel for $\alpha$-harmonic functions
defined by

$P_{{\alpha}}(z):=P_{0,{\alpha}}(z)=\frac{(1-|z|^{2})^{\alpha+1}}{(1-z)(1-\overline{z})^{\alpha+1}}.$

The aim of this paper is extend the previous results for
$(\alpha,\beta)$-harmonic functions for ${\alpha},{\beta}$ real numbers such
that ${\alpha},{\beta}\in{\mathbb{R}}\setminus{\mathbb{Z}}^{-}$ with
${\alpha}+{\beta}>-1$.

##  2\. Main results

Let us recall the polar partial derivative of $u$ with respect to $\theta$ is
given by

$\partial_{\theta}u=i(z\partial_{z}u-\overline{z}\partial_{\overline{z}}u).$
(2.1)

We start this section by the estimate of the angular derivative for
$({\alpha},{\beta})$-harmonic functions.

######  Theorem 2.1.

Assume that ${\alpha}+{\beta}>-1$ and $p\in[1,\infty]$. Let
$u=P_{{\alpha},{\beta}}[f]$ be an $({\alpha},{\beta})$-harmonic mapping on
${\mathbb{D}}$ with the function $f$ being absolutely continuous with
$\dot{f}\in L^{p}({\mathbb{T}})$. Then

$\partial_{\theta}u=P_{{\alpha},{\beta}}[\dot{f}]\in
h_{{\alpha},{\beta}}^{p}({\mathbb{D}}),$

and

$\|\partial_{\theta}u\|_{p}\leq|c_{\alpha,\beta}|\,c_{{\alpha}+{\beta}+1}\,\|\dot{f}\|_{L^{p}},$
(2.2)

where the constants $c_{{\alpha},{\beta}}$ and $c_{{\alpha}+{\beta}+1}$ are
defined in (1.2) and (1.11) respectively.

Proof. By assumptions, we have

$u(z)=P_{{\alpha},{\beta}}[f](z)=\frac{c_{{\alpha},{\beta}}}{2\pi}\int_{0}^{2\pi}K_{{\alpha},{\beta}}(ze^{-it})\,f(e^{it})\,dt.$

Let $z=re^{i\theta}$, an easy computations show that

$\partial_{\theta}K_{{\alpha},{\beta}}(ze^{-it})=-\partial_{t}K_{{\alpha},{\beta}}(ze^{-it}).$

An integration by parts leads to

$\partial_{\theta}u(z)=P_{{\alpha},{\beta}}[\dot{f}](z),$

By Jensen’s inequality, we get

$|\partial_{\theta}u(z)|^{p}\leq(|c_{{\alpha},{\beta}}|\,I_{{\alpha}+{\beta}+1}(r))^{p-1}\,\frac{1}{2\pi}\int_{0}^{2\pi}|K_{{\alpha},{\beta}}(ze^{-it})||\dot{f}(e^{it})|^{p}\,dt.$

Using Fubini’s theorem, it yields

$\frac{1}{2\pi}\int_{0}^{2\pi}|\partial_{\theta}u(re^{i\theta})|^{p}d\theta\leq(|c_{{\alpha},{\beta}}|\,I_{{\alpha}+{\beta}+1}(r))^{p}\|\dot{f}\|_{L^{p}}^{p}.$

Thus, using Lemma 1.1, we have

$\|\partial_{\theta}u\|_{p}\leq|c_{{\alpha},{\beta}}|\,c_{{\alpha}+{\beta}+1}\|\dot{f}\|_{L^{p}}.$

For $p=\infty$, it follows that

$|\partial_{\theta}u(z)|\leq|c_{{\alpha},{\beta}}|\,|I_{{\alpha}+{\beta}+1}(r)\|\dot{f}\|_{L^{\infty}}\leq|c_{{\alpha},{\beta}}|c_{{\alpha}+{\beta}+1}\|\dot{f}\|_{L^{\infty}}.$

∎

######  Lemma 2.1.

$\partial_{z}K_{{\alpha},{\beta}}(z)=\frac{({\alpha}+1)(1-|z|^{2})^{{\alpha}+{\beta}}}{(1-z)^{{\alpha}+2}(1-\overline{z})^{{\beta}}}-\frac{{\beta}\,\overline{z}(1-|z|^{2})^{{\alpha}+{\beta}}}{(1-z)^{{\alpha}+1}(1-\overline{z})^{{\beta}+1}},$
(2.3)

and

$\partial_{\overline{z}}K_{{\alpha},{\beta}}(z)=\frac{({\beta}+1)(1-|z|^{2})^{{\alpha}+{\beta}}}{(1-z)^{{\alpha}}(1-\overline{z})^{{\beta}+2}}-\frac{{\alpha}\,z(1-|z|^{2})^{{\alpha}+{\beta}}}{(1-z)^{{\alpha}+1}(1-\overline{z})^{{\beta}+1}}.$
(2.4)

Proof. Differentiating the kernel $K_{{\alpha},q}$ with respect to $z$, we get

$\partial_{z}K_{{\alpha},{\beta}}(z)=\bigg{(}-({\alpha}+{\beta}+1)\frac{\overline{z}}{1-|z|^{2}}+({\alpha}+1)\frac{1}{1-z}\bigg{)}K_{{\alpha},{\beta}}(z),$

see for instance [15, Lemma 1.1]. Next, observe that

$-({\alpha}+{\beta}+1)\frac{\overline{z}}{1-|z|^{2}}+({\alpha}+1)\frac{1}{1-z}=\frac{({\alpha}+1)(1-\overline{z})}{(1-z)(1-|z|^{2})}-\frac{{\beta}\overline{z}}{(1-|z|^{2})},$

and the conclusion (2.3) follows immediately. By the symmetry property of the
kernel $K_{{\alpha},{\beta}}$, we get (2.4). ∎

######  Proposition 2.1.

Let $u=P_{{\alpha},{\beta}}[f]$ be an $({\alpha},{\beta})$-harmonic mapping on
${\mathbb{D}}$ with $f$ being absolutely continuous on ${\mathbb{T}}$. Then

$z\partial_{z}u=-ic_{{\alpha},{\beta}}\left(K_{{\alpha}.{\beta}-1}[\dot{f}]+i{\beta}\overline{z}\,K_{{\alpha}-1,{\beta}}[f_{1}]\right),$
(2.5)
$\overline{z}\partial_{\overline{z}}u=ic_{{\alpha},{\beta}}\left(K_{{\alpha}-1.{\beta}}[\dot{f}]-i{\alpha}zK_{{\alpha},{\beta}-1}[f_{1}]\right),$
(2.6)

where $f_{1}$ is defined on ${\mathbb{T}}$ by $f_{1}(e^{it})=e^{it}f(e^{it})$.

Proof. The partial derivative of $u$ with respect to $z$ is giving by

$\partial_{z}u(z)=\frac{1}{2\pi}\int_{0}^{2\pi}\partial_{z}P_{{\alpha},{\beta}}(ze^{-it})e^{-it}f(e^{it})dt.$

By (2.3), we have

$\partial_{z}K_{{\alpha},{\beta}}(z)=\frac{({\alpha}+1)(1-|z|^{2})^{{\alpha}+{\beta}}}{(1-z)^{{\alpha}+2}(1-\overline{z})^{{\beta}}}-\frac{{\beta}\,\overline{z}(1-|z|^{2})^{{\alpha}+{\beta}}}{(1-z)^{{\alpha}+1}(1-\overline{z})^{{\beta}+1}}.$

Hence

$\displaystyle\frac{1}{c_{{\alpha},{\beta}}}\partial_{z}u(z)$ $\displaystyle=$
$\displaystyle\frac{1}{2\pi}(1-|z|^{2})^{{\alpha}+{\beta}}\int_{0}^{2\pi}\frac{({\alpha}+1)f(e^{it})e^{-it}}{(1-ze^{-it})^{{\alpha}+2}(1-\overline{z}e^{it})^{{\beta}}}\,dt$
$\displaystyle-$
$\displaystyle\frac{1}{2\pi}(1-|z|^{2})^{{\alpha}+{\beta}}\int_{0}^{2\pi}\frac{{\beta}\overline{z}f(e^{it})}{(1-ze^{-it})^{{\alpha}+1}(1-\overline{z}e^{it})^{{\beta}+1}}\,dt.$

Introduce the functions $h_{1}$ and $h_{2}$ defined on ${\mathbb{T}}$ by

$h_{1}(e^{it}):=\frac{({\alpha}+1)e^{-it}}{(1-ze^{-it})^{{\alpha}+2}(1-\overline{z}e^{it})^{{\beta}}}$

and

$h_{2}(e^{it}):=\frac{{\beta}\overline{z}}{(1-ze^{-it})^{{\alpha}+1}(1-\overline{z}e^{it})^{{\beta}+1}},$

so that we get the following expression

$\frac{1}{c_{{\alpha},{\beta}}}\partial_{z}u(z)=\frac{1}{2\pi}(1-|z|^{2})^{{\alpha}+{\beta}}\bigg{(}\int_{0}^{2\pi}h_{1}(e^{it})f(e^{it})dt-\int_{0}^{2\pi}h_{2}(e^{it})f(e^{it})dt\bigg{)}.$
(2.7)

Consider the function $g$ on ${\mathbb{T}}$ defined by

$g(e^{it}):=\frac{1}{(1-ze^{-it})^{{\alpha}+1}(1-\overline{z}e^{it})^{{\beta}}},$

its derivative $\dot{g}$ is given by

$\dot{g}(e^{it})=\frac{-ize^{-it}({\alpha}+1)}{(1-ze^{-it})^{{\alpha}+2}(1-\overline{z}e^{it})^{{\beta}}}+\frac{i\overline{z}{\beta}e^{it}}{(1-ze^{-it})^{{\alpha}+1}(1-\overline{z}e^{it})^{{\beta}+1}},$

that is

$\dot{g}(e^{it})\frac{}{}=-izh_{1}(e^{it})+ie^{it}h_{2}(e^{it}).$ (2.8)

Using (2.7) and (2.8), we have

$\displaystyle\frac{iz\partial_{z}u(z)}{c_{{\alpha},{\beta}}(1-|z|^{2})^{{\alpha}+{\beta}}}$
$\displaystyle=$
$\displaystyle\frac{1}{2\pi}\int_{0}^{2\pi}\big{(}-\dot{g}(e^{it})+ie^{it}h_{2}(e^{it})\big{)}f(e^{it})\,dt-\frac{1}{2\pi}\int_{0}^{2\pi}izh_{2}(e^{it})f(e^{it})\,dt$
$\displaystyle=$
$\displaystyle-\frac{1}{2\pi}\int_{0}^{2\pi}\dot{g}(e^{it})f(e^{it})dt+\frac{i}{2\pi}\int_{0}^{2\pi}\big{(}e^{it}-z\big{)}h_{2}(e^{it})f(e^{it})dt$

By integration by parts in the first integral and using the fact that
$e^{it}-z=e^{it}(1-ze^{-it})$, we obtain

$\displaystyle\frac{iz\partial_{z}u(z)}{c_{{\alpha},{\beta}}(1-|z|^{2})^{{\alpha}+{\beta}}}$
$\displaystyle=$
$\displaystyle\frac{1}{2\pi}\int_{0}^{2\pi}\frac{\dot{f}(e^{it})}{(1-ze^{-it})^{{\alpha}+1}(1-\overline{z}e^{it})^{{\beta}}}\,dt$
$\displaystyle+$
$\displaystyle\frac{i}{2\pi}\int_{0}^{2\pi}\frac{{\beta}\overline{z}e^{it}f(e^{it})}{(1-ze^{-it})^{{\alpha}}(1-\overline{z}e^{it})^{{\beta}+1}}\,dt.$

Finally, we deduce that

$iz\partial_{z}u=c_{{\alpha},{\beta}}\left(K_{{\alpha}.{\beta}-1}[\dot{f}]+i{\beta}\overline{z}K_{{\alpha}-1,{\beta}}[f_{1}]\right).$

The equation (2.6) is obtained by considering $\overline{u}$ which is
$({\beta},{\alpha})$-harmonic. ∎

Now, we are ready to state our first main result.

######  Theorem 2.2.

Assume that ${\alpha}+{\beta}>-1$ and $1\leq p\leq\infty$. Let
$u=P_{{\alpha},{\beta}}[f]$ be an $({\alpha},{\beta})$-harmonic mapping on
${\mathbb{D}}$ with $f$ being absolutely continuous and $\dot{f}\in
L^{p}({\mathbb{T}})$. Then

$M_{p}(r,z\partial_{z}u)\leq|c_{{\alpha},{\beta}}|\,I_{{\alpha}+{\beta}}(|z|)\bigg{(}\|\dot{f}\|_{L^{p}}+|{\beta}|\|f\|_{L^{p}}\bigg{)},$
(2.9)

and

$M_{p}(r,\overline{z}\partial_{\overline{z}}u)\leq|c_{{\alpha},{\beta}}|\,I_{{\alpha}+{\beta}}(|z|)\bigg{(}\|\dot{f}\|_{L^{p}}+|{\alpha}|\|f\|_{L^{p}}\bigg{)}.$
(2.10)

In particular

  1. (i)

if ${\alpha}+{\beta}>0$, then $\partial_{z}u$ and $\partial_{\overline{z}}u$
belong to $H_{G}^{p}({\mathbb{D}})$ and

$\|z\partial_{z}u\|_{p}\leq|c_{{\alpha},{\beta}}|\,c_{{\alpha}+{\beta}}\big{(}\|\dot{f}\|_{L^{p}}+|{\beta}|\|f\|_{L^{p}}\big{)},$
(2.11)

and

$\|\overline{z}\partial_{\overline{z}}u\|_{p}\leq|c_{{\alpha},{\beta}}|\,c_{{\alpha}+{\beta}}\big{(}\|\dot{f}\|_{L^{p}}+|{\alpha}|\|f\|_{L^{p}}\big{)}.$
(2.12)

  2. (ii)

If ${\alpha}+{\beta}<0$ and $p<-\frac{1}{{\alpha}+{\beta}}$, then
$\partial_{z}u$ and $\partial_{\overline{z}}u$ are in $L^{p}({\mathbb{D}})$.

  3. (iii)

If ${\alpha}+{\beta}<0$ and $p\geq-\frac{1}{{\alpha}+{\beta}}$, then there
exists $u$ an $({\alpha},{\beta})$-harmonic function such that $\partial_{z}u$
and $\partial_{\overline{z}}u$ $\not\in L^{p}({\mathbb{D}})$. Moreover,
$\partial_{z}u$ and
$\partial_{\overline{z}}u\not\in\mathcal{H}_{\mathcal{G}}^{1}({\mathbb{D}})$.

Proof. Using (2.5), we obtain

$M_{p}(r,z\partial_{z}u)\leq|c_{{\alpha},{\beta}}|\left(M_{p}(r,K_{{\alpha}.{\beta}-1}[\dot{f}])+|{\beta}||z|M_{p}(r,K_{{\alpha}-1,{\beta}}[f_{1}])\right).$

Similarly, as in the proof of Theorem 2.1, using Jensen’s inequality and
Fubini’s theorem, we have

$M_{p}(r,K_{{\alpha}.{\beta}-1}[\dot{f}])\leq
I_{{\alpha}+{\beta}}(|z|)\|\dot{f}\|_{L^{p}}\quad\mbox{and}\quad
M_{p}(r,K_{{\alpha}-1.{\beta}}[f_{1}])\leq
I_{{\alpha}+{\beta}}(|z|)\|f\|_{L^{p}}.$

Therefore, we obtain the desired result (2.9). The equations (2.11), (2.12)
and the assertion (ii) are immediate consequences of (2.9), (2.10) and Lemma
1.1.

(iii) Let $k\in{\mathbb{Z}}$ and

$M_{{\alpha},{\beta},k}(z):=P_{{\alpha},{\beta}}[e^{ik\theta}](z)=\frac{1}{2\pi}\int_{0}^{2\pi}P_{{\alpha},{\beta}}(ze^{-i\theta})e^{ik\theta}d\theta,\quad
z\in{\mathbb{D}}.$

From the identity, see for instance [15]

$\displaystyle\frac{(1-|z|^{2})^{{\alpha}+{\beta}+1}}{(1-z)^{{\alpha}+1}(1-{\bar{z}})^{{\beta}+1}}$
$\displaystyle=\sum_{m=0}^{\infty}\frac{({\alpha}+1)_{m}}{m!}F(-{\alpha},m-{\beta};m+1;|z|^{2})z^{m}$
$\displaystyle\quad\
+\sum_{m=1}^{\infty}\frac{({\beta}+1)_{m}}{m!}F(-{\beta},m-{\alpha};m+1;|z|^{2}){\bar{z}}^{m},$

it yields for $k\in\mathbb{N}$

$M_{{\alpha},{\beta},k}(z)=c_{{\alpha},{\beta}}\frac{({\alpha}+1)_{k}}{k!}\,F(-{\alpha},k-{\beta};k+1;|z|^{2})z^{k}.$

Differentiate $M_{{\alpha},{\beta},k}$ with respect to $\overline{z}$, using
(1.5) and (1.4), we obtain

$\displaystyle\partial_{\bar{z}}M_{{\alpha},{\beta},k}(z)$ $\displaystyle=$
$\displaystyle-
c_{{\alpha},{\beta}}\frac{(k-{\beta})({\alpha})_{k+1}}{(k+1)!}\,F(1-{\alpha},1+k-{\beta};k+2;|z|^{2})z^{k+1}$
$\displaystyle=$ $\displaystyle-
c_{{\alpha},{\beta}}\frac{(k-{\beta})({\alpha})_{k+1}}{(k+1)!}\,(1-|z|^{2})^{{\alpha}+{\beta}}F(k+{\alpha}+1,{\beta}+1;k+2;|z|^{2})z^{k+1}.$

By (1.3), remark that

$\lim_{|z|\to
1}F(k+{\alpha}+1,1+{\beta};k+2;|z|^{2})=\frac{\Gamma(k+2)\Gamma(-{\alpha}-{\beta})}{\Gamma(1-{\alpha})\Gamma(k+1-{\beta})}.$

Next, we will pick $k\not={\beta}$ to ensure that $\lim_{|z|\to
1}F(k+{\alpha}+1,1+{\beta};k+2;|z|^{2})$ is non-zero, that is, $1-{\alpha}$
and $k+1-{\beta}$ are not negative integers. In that case, we have

$|\partial_{\bar{z}}M_{{\alpha},{\beta},k}(z)|\approx(1-|z|^{2})^{{\alpha}+{\beta}}\quad(|z|\to
1^{-}).$

Thus $\partial_{\bar{z}}M_{{\alpha},{\beta},k}\not\in L^{p}({\mathbb{D}})$ for
$p\geq-\frac{1}{{\alpha}+{\beta}}$. In addition, since ${\alpha}+{\beta}<0$,
we have $\partial_{\bar{z}}M_{{\alpha},{\beta},k}\not\in
H^{1}_{G}({\mathbb{D}})$.

For the choice of $k$, we will consider the following three cases:

  1. (i)

For ${\beta}\in\mathbb{N}$, take $k={\beta}+1$. We have
$1-{\alpha}>1+{\beta}\geq 1$, we see that $\lim_{|z|\to
1}F({\alpha}+{\beta}+1,{\beta};{\beta}+2;|z|^{2})$ is non-zero.

  2. (ii)

The case ${\alpha}\in\mathbb{N}$ is treated similarly.

  3. (iii)

For both ${\alpha}$ and ${\beta}$ are non integers, take $k=0$. Trivially, as
${\beta}\not\in\mathbb{N}$, we have $1-{\beta}\not\in{\mathbb{Z}}^{-}$.

∎

These examples are introduced in [14] to show to failure of the stability
property of being Lipschitz continuous for $({\alpha},{\beta})$-harmonic
functions when ${\alpha}+{\beta}\in(-1,0)$.

###  2.1. Case $-1<{\alpha}+{\beta}<0$

In this section, we give necessary conditions on $u$ an
$({\alpha},{\beta})$-harmonic such that $\partial_{z}u$ or
$\partial_{\overline{z}}u$ is in $H_{G}^{1}({\mathbb{D}})$. We show that $u$
is zero or $u$ is polyharmonic.

As $u$ is $({\alpha},{\beta})$-harmonic, we consider its series expansion, see
(1.8)

$u(z)=\sum_{k\geq 0}c_{k}F(-{\alpha},k-{\beta};k+1;|z|^{2})z^{k}+\sum_{k\geq
1}c_{-k}F(-{\beta},k-{\alpha};k+1;|z|^{2})\overline{z}^{k}.$

For $n\in{\mathbb{Z}}$, let

$u_{n}(z):=\frac{1}{2\pi}\int_{0}^{2\pi}u(ze^{i\theta})e^{-in\theta}d\theta.$
(2.13)

Thus

$u_{-n}(z)=c_{-n}F(-{\beta},n-{\alpha};n+1;|z|^{2})\overline{z}^{n},\quad\mbox{
for }n\geq 1,$ (2.14)

and

$u_{n}(z)=c_{n}F(-{\alpha},n-{\beta};n+1;|z|^{2})z^{n},\quad\mbox{ for }n\geq
0.$ (2.15)

By differentiation under the integral sign in (2.13), we obtain

$\partial_{z}u_{n}(z)=\frac{1}{2\pi}\int_{0}^{2\pi}\partial_{z}u(ze^{i\theta})e^{i(1-n)\theta}d\theta,$
(2.16)

$\partial_{\overline{z}}u_{n}(z)=\frac{1}{2\pi}\int_{0}^{2\pi}\partial_{\overline{z}}u(ze^{i\theta})e^{-i(1+n)\theta}d\theta.$

Differentiating the equations (2.14) and (2.15) with respect to $z$, and using
(1.4), we get

$\partial_{z}u_{-n}(z)=c_{-n}\frac{(-{\beta})(n-{\alpha})}{n+1}(1-|z|^{2})^{{\alpha}+{\beta}}F(n+{\beta}+1,{\alpha}+1;n+2;|z|^{2})\overline{z}^{n+1},$
(2.17)

$\displaystyle\partial_{z}u_{n}(z)$ $\displaystyle=$ $\displaystyle
c_{n}\frac{(-{\alpha})(n-{\beta})}{n+1}(1-|z|^{2})^{{\alpha}+{\beta}}F(n+{\alpha}+1,{\beta}+1;n+2;|z|^{2})\overline{z}z^{n}$
(2.18) $\displaystyle+$ $\displaystyle
nc_{n}F(-{\alpha},n-{\beta};n+1,|z|^{2})z^{n-1}.$

######  Theorem 2.3.

Let ${\alpha}$ and ${\beta}$ be non-integers such that $-1<{\alpha}+{\beta}<0$
and let $u$ be an $({\alpha},{\beta})$-harmonic function. If $\partial_{z}u$
or $\partial_{\overline{z}}u$ is in $H_{G}^{1}({\mathbb{D}})$, then $u$ is
zero.

Proof. Assume that $\partial_{z}u\in H_{G}^{1}({\mathbb{D}})$. Using (2.16),
we conclude that $\partial_{z}u_{n}$ is bounded on ${\mathbb{D}}$ for all
$n\in{\mathbb{Z}}$.
On one hand, according to (2.17), for $n\geq 1$, we have

$\partial_{z}u_{-n}(z)=c_{-n}\frac{(-{\beta})(n-{\alpha})}{n+1}(1-|z|^{2})^{{\alpha}+{\beta}}F(n+{\beta}+1,{\alpha}+1;n+2;|z|^{2})\overline{z}^{n+1},$

and by the property (1.3), we have

$\lim_{|z|\to
1}F(n+{\beta}+1,{\alpha}+1;n+2;|z|^{2})=\frac{\Gamma(n+2)\Gamma(-{\alpha}-{\beta})}{\Gamma(1-{\beta})\Gamma(n+1-{\alpha})}.$
(2.19)

As ${\alpha}$ and ${\beta}$ are non-integers, we deduce that the limit in
(2.19) is non-zero, therefore $\partial_{z}u_{-n}$ cannot be bounded unless
$c_{-n}=0$, for $n\geq 1$.

On the other hand, according to (2.18), the partial derivative with respect to
$z$ of $u_{n}$ is given by

$\displaystyle\partial_{z}u_{n}(z)$ $\displaystyle=$ $\displaystyle
c_{n}\frac{(-{\alpha})(n-{\beta})}{n+1}(1-|z|^{2})^{{\alpha}+{\beta}}F(n+{\alpha}+1,{\beta}+1;n+2;|z|^{2})\overline{z}z^{n}$
$\displaystyle+$ $\displaystyle
nc_{n}F(-{\alpha},n-{\beta};n+1,|z|^{2})z^{n-1}.$

As ${\alpha}+{\beta}>-1$, remark that the function
$F(-{\alpha},n-{\beta};n+1;|z|^{2})$ has a finite limit as $|z|$ goes to $1$,
therefore it is bounded on ${\mathbb{D}}$. In addition notice that
$\displaystyle\lim_{|z|\to 1}F(n+{\alpha}+1,{\beta}+1;n+2;|z|^{2})$ is non-
zero, as in (2.19). Therefore, since $\partial_{z}u_{n}$ is bounded, it yields
$c_{n}=0$, for $n\geq 0$. Finally, we conclude that $u=0$.
The case $\partial_{\overline{z}}u\in H_{G}^{1}({\mathbb{D}})$ is treated
similarly and we conclude that $u$ must be zero. ∎

######  Theorem 2.4.

Let ${\alpha}$ be a positive integer and ${\beta}$ be a real number such that
$-1<{\alpha}+{\beta}<0$. Let $u$ be an $({\alpha},{\beta})$-harmonic mapping.
If $\partial_{z}u$ or $\partial_{\overline{z}}u$ is in
$H_{G}^{1}({\mathbb{D}})$, then $c_{-n}=0$ for $n>{\alpha}$, where $c_{-n}$
are coefficients in the series expansion of $u$ in (1.8).
Indeed, $u$ is $({\alpha}+1)$-harmonic, and

$u(z)=K_{0}(z)+K_{1}(z)|z|^{2}+\ldots+K_{{\alpha}}(z)|z|^{2{\alpha}},$

where $K_{i}=h_{i}+\overline{g_{i}}$ for $i=0,\ldots,{\alpha}-1$ and
$K_{{\alpha}}$ is analytic, and $g_{i}$ are polynomials and $h_{i}$ are
analytic for $i=0,\ldots,{\alpha}-1$.

Let $n$ be a positive integer, recall that a function $u$ is called
$n$-harmonic if $\Delta^{n}u=0$.

Proof.

(a) Assume that $\partial_{z}u\in H^{1}_{G}({\mathbb{D}})$, then we deduce
from (2.16) that $\partial_{z}u_{n}$ is bounded for all $n\in{\mathbb{Z}}$.
Using (2.17), we see that the limit in (2.19) cannot be zero for $n>{\alpha}$
as by assumptions ${\beta}$ is non zero and $1-{\beta}$ cannot be zero or a
negative integer. Hence we conclude that $c_{-n}=0$ for $n>{\alpha}$.
Therefore

$u(z)=\sum_{k\geq
0}c_{k}F(-{\alpha},k-{\beta};k+1;|z|^{2})z^{k}+\sum_{k=1}^{\alpha}c_{-k}F(-{\beta},k-{\alpha};k+1;|z|^{2})\overline{z}^{k}.$

We should notice that the functions $F(-{\alpha},k-{\beta};k+1;x)$ and
$F(-{\beta},k-{\alpha};k+1;x)$ are polynomials respectively of degree
${\alpha}$ and ${\alpha}-k$ ($1\leq k\leq{\alpha}$), and we have

$\sum_{k\geq
0}c_{k}F(-{\alpha},k-{\beta};k+1;|z|^{2})z^{k}=h_{0}(z)+h_{1}(z)|z|^{2}+\ldots+h_{{\alpha}}(z)|z|^{2{\alpha}},$

where $h_{i}$ are analytic for $i=0,\ldots,{\alpha}$. In addition,

$\sum_{k=1}^{\alpha}c_{-k}F(-{\beta},k-{\alpha};k+1;|z|^{2})\overline{z}^{k}=\overline{g_{0}(z)}+\overline{g_{1}(z)}|z|^{2}+\ldots+\overline{g_{{\alpha}-1}(z)}\,|z|^{2({\alpha}-1)},$

where $g_{i}$ are polynomials for $i=0,\ldots,{\alpha}-1$. Therefore, we
obtain the desired result.

(b) Assume that $\partial_{\overline{z}}u\in H^{1}_{G}({\mathbb{D}})$. As in
the previous case, we deduce that $\partial_{\overline{z}}u_{-n}$ is bounded
for $n\geq 1$.

$\displaystyle\partial_{\overline{z}}u_{-n}(z)$ $\displaystyle=$
$\displaystyle
c_{-n}\frac{{\beta}({\alpha}-n)}{n+1}(1-|z|^{2})^{{\alpha}+{\beta}}F(n+{\beta}+1,1+{\alpha};n+2;|z|^{2})\overline{z}^{n}z$
$\displaystyle+$ $\displaystyle
nc_{-n}F(-{\beta},n-{\alpha};n+1;|z|^{2})\overline{z}^{n-1}.$

As ${\alpha}+{\beta}>-1$, remark that the function
$F(-{\beta},n-{\alpha};n+1;|z|^{2})$ has a finite limit as $|z|$ goes to $1$,
therefore it is bounded on ${\mathbb{D}}$. Similarly as in the previous case
$\displaystyle\lim_{|z|\to 1}F(n+{\beta}+1,{\alpha}+1;n+2;|z|^{2})$ is non-
zero, for $n>{\alpha}$. Therefore $c_{-n}=0$, for $n>{\alpha}$. ∎

For the converse, we prove the following

######  Theorem 2.5.

Let ${\alpha}$ be a positive integer and ${\beta}$ be a real number such that
${\alpha}+{\beta}\in(-1,0)$. Assume
$f\in\mathcal{C}^{{\alpha}-1}({\mathbb{T}})$ with $f^{({\alpha}-1)}$
absolutely continuous and $f^{({\alpha})}\in L^{p}({\mathbb{T}})$ for some
$p\in(1,\infty)$ and $u=P_{{\alpha},{\beta}}[f]$ with $c_{-n}=0$ for
$n>{\alpha}$. Then $\partial_{z}u$ and $\partial_{\overline{z}}u$ are in
$H^{p}_{G}({\mathbb{D}})$.

Here $f^{(0)}=f$ and
$f^{({\alpha})}(e^{i\theta}):=\frac{d}{d\theta}f^{({\alpha}-1)}(e^{i\theta})$.
Proof. As $c_{-n}=0$ for $n>{\alpha}$, $u$ can be written as

$u(z)=\sum_{n\geq
0}c_{n}F(-{\alpha},n-{\beta};n+1;|z|^{2})z^{n}+\sum_{n=1}^{\alpha}c_{-n}F(-{\beta},n-{\alpha};n+1;|z|^{2})\overline{z}^{n}.$

Remark that for $n\in[\\![1..{\alpha}]\\!]$, the hypergeometric function
$F(-{\beta},n-{\alpha};n+1;|z|^{2})$ is a polynomial, therefore the mapping
$z\mapsto\sum_{n=1}^{\alpha}c_{-n}F(-{\beta},n-{\alpha};n+1;|z|^{2})\overline{z}^{n}$
is smooth on $\overline{{\mathbb{D}}}$. Therefore $\partial_{\overline{z}}u\in
H^{p}({\mathbb{D}})$ if and only $\partial_{\overline{z}}u_{1}\in
H^{p}({\mathbb{D}})$, where

$u_{1}(z)=\sum_{n\geq 0}c_{n}F(-{\alpha},n-{\beta};n+1;|z|^{2})z^{n}.$

Since

$F(-{\alpha},n-{\beta};n+1;|z|^{2})=\sum_{k=0}^{\alpha}\frac{(-{\alpha})_{k}(n-{\beta})_{k}}{(n+1)_{k}k!}|z|^{2k},$

we write

$u_{1}(z)=H_{0}(z)+H_{1}(z)|z|^{2}+\ldots+H_{{\alpha}}(z)|z|^{2{\alpha}},$

where for $k\in[\\![0..{\alpha}]\\!]$, $H_{k}$ is analytic and

$H_{k}(z)=\frac{(-{\alpha})_{k}}{k!}\sum_{n\geq
0}\frac{(n-{\beta})_{k}}{(n+1)_{k}}c_{n}z^{n}.$

In addition, by (1.9), we have

$c_{n}\,\frac{\Gamma(n+1)\Gamma({\alpha}+{\beta}+1)}{\Gamma({\beta}+1)\Gamma(n+{\alpha}+1)}=\hat{f}(n)\mbox{
for }n\geq 0.$

Thus, as

$\frac{(n+1)_{{\alpha}}}{(n+1)_{k}}=(n+k+1)_{{\alpha}-k}\mbox{ for
}k\in[\\![0..{\alpha}]\\!],$

we get

$H_{k}(z)=\frac{(-{\alpha})_{k}\Gamma({\beta}+1)}{\Gamma({\alpha}+{\beta}+1)k!}\sum_{n\geq
0}(n-{\beta})_{k}(n+1+k)_{{\alpha}-k}\,\hat{f}(n)z^{n},$

and $H_{k}$ is a linear combination of $Q_{\ell}(z):=\displaystyle\sum_{n\geq
0}(in)^{\ell}\hat{f}(n)z^{n}$ for $\ell=0,\ldots,{\alpha}$.

We claim that

$Q_{\ell}\in\bigcap_{1<q<\infty}H^{q}({\mathbb{D}})\mbox{ for
}\ell=0\ldots{\alpha}-1\mbox{ and }Q_{\alpha}\in H^{p}({\mathbb{D}}).$

Indeed, consider the harmonic function defined as the Poisson extension of $f$
i.e.,

$U(z)=P[f]=\sum_{n\geq 0}\hat{f}(n)z^{n}+\sum_{n\geq
1}\hat{f}(-n)\overline{z}^{n}.$

For $\ell=1,\ldots,{\alpha}$, we have

$\frac{d^{\ell}}{d\theta^{\ell}}U(z)=P[f^{\ell}]=\sum_{n\geq
0}(in)^{\ell}\hat{f}(n)z^{n}+\sum_{n\geq
1}(-in)^{\ell}\hat{f}(-n)\overline{z}^{n}.$

Hence $Q_{\ell}=P_{+}(\frac{d^{\ell}U}{d\theta^{\ell}}),$ the Riesz projection
of the harmonic function $\frac{d^{\ell}U}{d\theta^{\ell}}$. Thus the claim
follows from Riesz’s theorem on conjugate functions. Since
$\partial_{\overline{z}}u_{1}=zH_{1}(z)+\ldots+{\alpha}z^{{\alpha}}H_{{\alpha}}(z)\overline{z}^{{\alpha}-1},$
this implies that $\partial_{\overline{z}}u\in H^{p}_{G}({\mathbb{D}})$.
On the other hand, as
$\partial_{\theta}u=i(z\partial_{z}u-\overline{z}\partial_{\overline{z}}u)\in
H^{p}_{G}({\mathbb{D}})$, we conclude that $\partial_{z}u\in
H^{p}_{G}({\mathbb{D}})$.∎

We expect a stronger version of Theorem 2.5 by relaxing the assumptions of the
boundary function $f$. For $\alpha=1$, it yields

######  Corollary 2.1.

Let $\beta\in(-2,-1)$ and $p\in(1,\infty)$ and $u=P_{1,\beta}[f]$ where $f$ is
absolutely continuous on ${\mathbb{T}}$ with $\dot{f}\in L^{p}({\mathbb{T}})$.
Then $\partial_{z}u$ or $\partial_{\overline{z}}u$ is in
$H_{G}^{1}({\mathbb{D}})$, if and only if $\hat{f}(-n)=0$ for $n>1$. In
particular $u$ is biharmonic.

###  2.2. Case ${\alpha}+{\beta}=0$

######  Theorem 2.6.

Let ${\alpha},{\beta}$ be non zero numbers in
${\mathbb{R}}\setminus{\mathbb{Z}}^{-}$ such that ${\alpha}+{\beta}=0$ and let
$u$ be a $({\alpha},{\beta})$-harmonic function. If $\partial_{z}u$ or
$\partial_{\overline{z}}u$ is in $H_{G}^{1}({\mathbb{D}})$, then $u$ is zero.

Proof. Assume that $\partial_{z}u\in H_{G}^{1}({\mathbb{D}})$. Using (2.13),
we conclude that $\partial_{z}u_{n}$ is bounded on ${\mathbb{D}}$. According
to (2.17), we have

$\partial_{z}u_{-n}(z)=c_{-n}\frac{{\alpha}(n-{\alpha})}{n+1}F(n-{\alpha}+1,{\alpha}+1;n+2;|z|^{2})\overline{z}^{n+1},$

and by (1.6), we have

$F(n-{\alpha}+1,{\alpha}+1;n+2;x)\approx\frac{\Gamma(n+2)}{\Gamma(n-{\alpha}+1)\Gamma({\alpha}+1)}\log(1-x),\quad(x\to
1^{-}).$ (2.20)

As ${\alpha}$ is not an integer, we deduce that the limit of the
hypergeometric function in (2.20) is $\infty$ as $x\to 1$, therefore
$\partial_{z}u_{-n}$ cannot be bounded unless $c_{-n}=0$, for $n\geq 1$.

Moreover, according to (2.18), the partial derivative with respect to $z$ of
$u_{n}$ is given by

$\displaystyle\partial_{z}u_{n}(z)$ $\displaystyle=$ $\displaystyle
c_{n}\frac{(-{\alpha})(n+{\alpha})}{n+1}F(n+{\alpha}+1,1-{\alpha};n+2;|z|^{2})\overline{z}z^{n}$
$\displaystyle+$ $\displaystyle
nc_{n}F(-{\alpha},n+{\alpha};n+1,|z|^{2})z^{n-1}.$

Remark that the function $F(-{\alpha},n+{\alpha};n+1;|z|^{2})$ has a finite
limit as $|z|$ goes to $1$, therefore it is bounded on ${\mathbb{D}}$. In
addition notice that, as in (2.20), $\lim_{|z|\to
1}F(n+{\alpha}+1,1-{\alpha};n+2;|z|^{2})=\infty$. Therefore
$\partial_{z}u_{n}$ is bounded unless $c_{n}=0$, for $n\geq 0$. Finally, we
conclude that $u=0$. ∎

Finally, given $u=P_{{\alpha},{\beta}}[f]$ a quasi-regular
$({\alpha},{\beta})$-harmonic function with $\dot{f}\in L^{p}({\mathbb{T}})$,
we prove that both $\partial_{z}u$ and $\partial_{\overline{z}}u$ are in
$H_{G}^{p}({\mathbb{D}})$ for $p\in[1,\infty]$.

Recall that a mapping $u:{\mathbb{D}}\to{\mathbb{C}}$ is said to be $K-$quasi-
regular, if

$|\partial_{z}u(z)|+|\partial_{\overline{z}}u(z)|\leq
K(|\partial_{z}u(z)|-|\partial_{\overline{z}}u(z)|),$

on ${\mathbb{D}}$, where $K\in[1,\infty)$ is a constant.

######  Theorem 2.7.

Suppose $1\leq p\leq\infty$,
${\alpha},{\beta}\in{\mathbb{R}}\setminus{\mathbb{Z}}^{-}$ such that
${\alpha}+{\beta}>-1$ and $u=P_{{\alpha},{\beta}}[f]$ is an
$({\alpha},{\beta})$-harmonic quasi-regular mapping of ${\mathbb{D}}$ with the
boundary function $f$ is absolutely continuous and satisfies $\dot{f}\in
L^{p}({\mathbb{T}})$ Then $\partial_{z}u$ and $\partial_{\overline{z}}u$ are
in $H_{G}^{p}({\mathbb{D}})$.

Proof. Let $z=re^{i\theta}$, by using the identity (2.1) and the inequality
(2.2), we get

$\frac{1}{2\pi}\int_{0}^{2\pi}\big{(}|z\partial_{z}u(re^{i\theta})|-|\overline{z}\partial_{\overline{z}}u(re^{i\theta})|\big{)}^{p}d\theta\leq\frac{1}{2\pi}\int_{0}^{2\pi}|\partial_{\theta}u(re^{i\theta})|^{p}d\theta\leq(|c_{{\alpha},{\beta}}|\,c_{{\alpha}+{\beta}+1})^{p}\|\dot{f}\|^{p}_{L^{p}},$

where $1\leq p<\infty.$ As $u$ is quasi-regular, it follows that there exists
a constant $K\geq 1$, such that

$|\partial_{z}u(z)|+|\partial_{\overline{z}}u(z)|\leq
K\big{(}|\partial_{z}u(z)|-|\partial_{\overline{z}}u(z)|\big{)}.$

Hence

$\frac{1}{2\pi}\int_{0}^{2\pi}\big{(}|z\partial_{z}u(re^{i\theta})|+|\overline{z}\partial_{\overline{z}}u(re^{i\theta})|\big{)}^{p}d\theta\leq(K|c_{{\alpha},{\beta}}|c_{{\alpha}+{\beta}+1})^{p}\|\dot{f}\|^{p}_{L^{p}},$

it yields that

$\frac{1}{2\pi}\int_{0}^{2\pi}|z\partial_{z}u(re^{i\theta})|^{p}d\theta\leq(K|c_{{\alpha},{\beta}}|c_{{\alpha}+{\beta}+1})^{p}\|\dot{f}\|^{p}_{L^{p}}$

and

$\frac{1}{2\pi}\int_{0}^{2\pi}|\overline{z}\partial_{\overline{z}}u(re^{i\theta})|^{p}d\theta\leq\bigg{(}\frac{K-1}{2}|c_{{\alpha},{\beta}}|c_{{\alpha}+{\beta}+1}\bigg{)}^{p}\|\dot{f}\|^{p}_{L^{p}}.$

Thus

$M_{p}(r,z\partial_{z}u)\leq
K|c_{{\alpha},{\beta}}|c_{{\alpha}+{\beta}+1}\|\dot{f}\|_{L^{p}}\quad\quad\text{and}\quad\quad
M_{p}(r,\overline{z}\partial_{\overline{z}}u)\leq\frac{K-1}{2}c_{{\alpha}+{\beta}+1}|c_{{\alpha},{\beta}}|\|\dot{f}\|_{L^{p}}.$

Hence $\partial_{z}u$ and $\partial_{\overline{z}}u$ are in
$H_{G}^{p}({\mathbb{D}})$. ∎

##  3\. $T_{\alpha}$-harmonic functions

Let ${\alpha}\in{\mathbb{R}}$ and denote by

$\omega_{\alpha}(z)=(1-|z|^{2})^{\alpha};\quad z\in{\mathbb{D}},$

the corresponding standard weight in the unit disc ${\mathbb{D}}$. The
associated weighted Laplace operators are given by

$L_{\alpha}=\partial_{z}\omega_{{\alpha}}^{-1}\partial_{\overline{z}}\quad\mbox{
and
}L_{\alpha}^{*}=\partial_{\overline{z}}\omega_{{\alpha}}^{-1}\partial_{z}\mbox{
in }{\mathbb{D}}.$

One can check that

$L_{\alpha}={\alpha}(1-|z|^{2})^{-{\alpha}-1}\overline{z}\partial_{\overline{z}}+(1-|z|^{2})^{\alpha}\partial_{z}\partial_{\overline{z}}.$

Let

$T_{\alpha}:=-\frac{{\alpha}^{2}}{4}\omega_{{\alpha}+1}^{-1}+\frac{1}{2}(L_{{\alpha}}+L^{*}_{{\alpha}}),$

that is,

$T_{\alpha}=-\frac{{\alpha}^{2}}{4}(1-|z|^{2})^{-{\alpha}-1}+\frac{{\alpha}}{2}(1-|z|^{2})^{-{\alpha}-1}\overline{z}\partial_{\overline{z}}+\frac{{\alpha}}{2}(1-|z|^{2})^{-{\alpha}-1}z\partial_{z}+(1-|z|^{2})^{-{\alpha}}\partial_{z}\partial_{\overline{z}}.$

Thus

$T_{\alpha}=(1-|z|^{2})^{-{\alpha}-2}\Delta_{{\alpha}/2,{\alpha}/2}.$

For ${\alpha}>-1$, we denote by

$M_{\alpha}:=M_{{\alpha}/2,{\alpha}/2}=P_{{\alpha}/2,{\alpha}/2}[1].$

Recall that a function $u$ is $T_{\alpha}$-harmonic if $T_{\alpha}u=0$. Remark
that $T_{\alpha}$-harmonic functions are exactly
$({\alpha}/2,{\alpha}/2)$-harmonic functions. Thus combining Theorem 2.2 and
Theorem 2.3, we obtain

######  Corollary 3.1.

Let $\alpha>-1$ and $f$ be an absolutely continuous such that $\dot{f}\in
L^{p}({\mathbb{T}})$ with $1\leq p\leq\infty$ and
$u=P_{{\alpha}/2,{\alpha}/2}[f]$. Then

  1. (i)

If ${\alpha}>0$, then $\partial_{z}u$ and $\partial_{\overline{z}}u$ are in
$H_{G}^{p}({\mathbb{D}})$.

  2. (ii)

If ${\alpha}<0$, then $\partial_{z}u$ or $\partial_{\overline{z}}u$ in
$H_{G}^{1}({\mathbb{D}})$, if and only if, $u$ is zero.

  3. (iii)

If ${\alpha}<0$ and $p<-\frac{1}{{\alpha}}$, then $\partial_{z}u$ and
$\partial_{\overline{z}}u$ are in $L^{p}({\mathbb{D}})$.

  4. (iv)

If ${\alpha}<0$ and $p\geq-\frac{1}{{\alpha}}$, then
$\partial_{z}M_{{\alpha}}$ and $\partial_{\overline{z}}M_{\alpha}$ $\not\in
L^{p}({\mathbb{D}})$.

Our results simplify those obtained by Chen et al. in [6].

## References

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properly extends $A$; we must prove that $B$ embeds in $M$ over $A$. One
reduces to the case where $B$ is transformally algebraic and
$\operatorname{acl}_{\omega}$-finitely generated over $A$. In the case where
$A$ is transformally Newtonian, using Lemma 9.11 below, one can find
$A\subseteq C\subseteq B$ with $\operatorname{acl}_{\omega}C=C$ which properly
extends $A$ and such that $C$ embeds in $M$ over $A$. Now the field $C$ is not
in general transformally Newtonian; in this case we replace $C$ with a
transformally Newtonian model $C^{\star}$ inside $M$, replace $B$ with
$B^{\star}=\operatorname{acl}_{\omega}\left(B\otimes_{C}C^{\star}\right)$ and
repeat the process.

One can show that in retrospect that this process terminates in finitely many
stages, for example by induction on SU rank or the inertial dimension of $B$
over $A$; in what follows we give a soft proof in which the process is
repeated transfinitely, exploiting the fact that the valuation group does not
increase in transformally algebraic extensions. An alternative approach
directly using scatteredness is given in Remark 9.12

Before we turn to the proof of Theorem 9.10 we state and prove the Lemma
promised in the informal description:

######  Lemma 9.11.

Let $M$ be a saturated model of $\widetilde{\operatorname{VFA}}$. Let
$A\subseteq M$ be countable and transformally Newtonian. Let
$B=\operatorname{acl}_{\omega}\left(Ab\right)$ be a transformally algebraic
extension of $A$ which is algebraically closed and transformally Henselian,
where $b\in B$ is a single element lying outside $A$. Then there is an element
$c\in B$ with the following properties:

(1) We have the inequality $v\left(b-c\right)>v\left(a-c\right)$ for all $a\in
A$

(2) The field $\operatorname{acl}_{\omega}\left(Ac\right)$ embeds in $M$ over
$A$

###### Proof.

We first reduce ourselves to the case where $vb=0$ and the residue class of
$b$ is not in $k_{A}$. Since $A$ is transformally Newtonian, it is
transformally algebraically maximal and we have
$\Gamma_{A}=\Gamma_{A}\otimes\mathbf{Q}\left(\sigma\right)$. So as $B$ is
transformally algebraic over $A$, there is an element $a^{\star}\in A$
maximizing the quantity $v\left(b-a^{\star}\right)$, i.e, we have
$v\left(b-a^{\star}\right)\geq v\left(b-a\right)$ for all $a\in A$. Let
$b^{\star}=b-a^{\star}$. Replacing $b$ by $b^{\star}$ is harmless: if
$c^{\star}$ witnesses the truth of the statement for $b^{\star}$, then
$c=c^{\star}+a^{\star}$ witnesses the truth of the original statement. We may
therefore assume $a^{\star}=0$ maximizes the valuative distance, i.e we have
the inequality $v\left(b\right)\geq v\left(b-a\right)$ for all $a\in A$. In a
similar fashion, using
$\Gamma_{A}=\Gamma_{A}\otimes\mathbf{Q}\left(\sigma\right)=\Gamma_{B}$, after
replacing $b$ by a scalar multiple of an element of $A$, we may assume that
$vb=0$; so the residue class of $b$ is not in $k_{A}$.

Let $A^{\text{tunr}}$ be the maximal transformally unramified extension of $A$
in $B$, so $k_{A^{\text{tunr}}}=k_{B}$; so the field $A^{\text{tunr}}$ is the
universal transformally Henselian extension of $A$ reproducing the embedding
$k_{A}\hookrightarrow k_{B}$ at the level of residue fields. Since $M$ is a
model of $\widetilde{\operatorname{VFA}}$, it is strictly transformally
Henselian, so the field $A^{\text{tunr}}$ embeds in $M$ over $A$. Since the
fields $A$ and the residue field $k_{A^{\text{tunr}}}$ is algebraically
closed, it follows from 5.2 that $A^{\text{tunr}}$ admits no nontrivial finite
$\sigma$-invariant Galois extensions; using $M=\operatorname{acl}_{\omega}M$
and Corollary 5.4, we can further find an embedding of
$\operatorname{acl}_{\omega}\left(A^{\text{tunr}}\right)$ in $M$ over $A$.
Finally, since $k_{A^{\text{tunr}}}=k_{B}$ and the residue class of $b$ is not
in $k_{A}$, we can find an element
$c\in\operatorname{acl}_{\omega}\left(A^{\text{tunr}}\right)$ which is as in
the lemma (namely any element whose residue class coincides with that of $b$).
∎

Proof of Theorem 9.10 We now turn to the details.

Step 1. Every model of $\operatorname{VFA}$ embeds in a model of
$\widetilde{\operatorname{VFA}}$. Let $K$ be a model of $\operatorname{VFA}$.
If $K$ is trivially valued then we may embed it in a field of Hahn series so
assume that $K$ is nontrivially valued. Using Lemma 6.11 we may assume that
$\Gamma=\Gamma\otimes\mathbf{Q}\left(\sigma\right)$ and using Proposition 5.29
we may assume that $K$ is strictly transformally Henselian. By Corollary 6.28
we may assume that $K$ is transformally Newtonian; this implies that every
nonconstant difference polynomial with coefficients in $K$ has a $K$-root; as
this holds in particular for additive operators, we find that $K$ is a model
of $\widetilde{\operatorname{VFA}}$

Next we show that the models of $\widetilde{\operatorname{VFA}}$ are
existentially closed. Fix a saturated model $M$ of
$\widetilde{\operatorname{VFA}}$. Let $A\subseteq M$ be countable with
$\operatorname{acl}_{\omega}A=A$. Let $B=\operatorname{acl}_{\omega}B$ be a
countable extension of $A$; we want to prove that $B$ embeds in $M$ over $A$.

Step 2. We first deal with the case where $B$ is transformally algebraic over
$A$. By compactness, we may assume that $B$ is
$\operatorname{acl}_{\omega}$-finitely generated over $A$, that is, we have
$B=\operatorname{acl}_{\omega}\left(Ab\right)$ for a finite tuple $b\in B$. By
induction on the number of generators, we may moreover assume that $b$ is a
single element.

We shall construct continuous transfinite sequences

$A=A_{0}\subseteq A_{1}\subseteq A_{2}\subseteq\ldots\subseteq
A_{\alpha}\subseteq\ldots$

and

$B=B_{0}\subseteq B_{1}\subseteq\ldots\subseteq B_{\alpha}\subseteq\ldots$

of models of $\operatorname{VFA}$ with
$B_{\alpha}=\operatorname{acl}_{\omega}B_{\alpha}$ and
$A_{\alpha}=\operatorname{acl}_{\omega}A_{\alpha}$, for each countable ordinal
$\alpha$. Each $A_{\alpha}$ will be equipped with an embedding
$i_{\alpha}\colon A_{\alpha}\hookrightarrow M$ and an embedding
$j_{\alpha}\colon A_{\alpha}\hookrightarrow B_{\alpha}$ over $A=A_{0}$. These
embeddings will be coherent in the sense that for $\alpha<\beta$ we have that
$i_{\alpha}$ is the restriction of $i_{\alpha}$ to $A_{\alpha}$ and likewise
for $j_{\alpha}$ and $j_{\beta}$. Furthermore, the fields $B_{\alpha}$ will
all be transformally algebraic over $B=B_{0}$. Finally, for each $\alpha$, if
$b\notin A_{\alpha}$, then there will be some $\beta>\alpha$ and some
$c_{\beta}\in A_{\beta}$ with the property that
$v\left(b-c_{\beta}\right)>v\left(b-a\right)$ for all $a\in A_{\alpha}$. Since
$\Gamma_{B_{\alpha}}=\Gamma_{A}\otimes\mathbf{Q}\left(\sigma\right)$ such a
sequence must eventually terminate; the embedding of $A_{\alpha}$ in $M$ over
$A$ for $\alpha$ large will thus restrict to an embedding of
$\operatorname{acl}_{\omega}\left(Ab\right)$ in $M$ over $A$, as wanted.

The construction is as follows. For $\alpha$ a limit ordinal take the union.
If $\alpha$ is even, let $A_{\alpha+1}$ be a countable transformally Newtonian
model of $\operatorname{VFA}$ over $A_{\alpha}$ inside $M$, and let
$B_{\alpha+1}=\operatorname{acl}_{\omega}\left(B_{\alpha}\otimes_{A_{\alpha}}A_{\alpha+1}\right)$;
thus $B_{\alpha+1}$ is an algebraically closed and transformally Henselian
extension of $A_{\alpha}$ containing copies of $B_{\alpha}$ and
$A_{\alpha+1}$, and generated as an algebraically closed and transformally
Henselian field by the images of those fields; the choice of $B_{\alpha+1}$ is
not in general unique, even up to isomorphism, but its existence is guaranteed
by Theorem 9.1. If $\alpha$ is odd, we let $B_{\alpha+1}=B_{\alpha}$ and
$A_{\alpha+1}=\operatorname{acl}_{\omega}\left(A_{\alpha}c_{\alpha+1}\right)$,
where $c_{\alpha+1}\in B_{\alpha+1}$ is the element provided by Lemma 9.11.

Step 3. For the general case, increasing $B$ (while keeping it countable) we
may assume that it is itself a model of $\widetilde{\operatorname{VFA}}$ and
in particular transformally Newtonian. Let $A\subseteq C\subseteq B$ be a
model of $\operatorname{VFA}$ with $C=\operatorname{acl}_{\omega}C$ maximal
with respect to the property that it embeds in $M$ over $A$. By Step 2, we
find that no element of $B$ is transformally algebraic over $C$ unless it
already lies in $C$. Since $B$ is a model of $\widetilde{\operatorname{VFA}}$,
it is transformally Newtonian, so as $C$ is transformally algebraically closed
in $B$, the field $C$ is transformally Newtonian also. By Corollary 6.30 we
find every $a\in B$ not in $C$ is generic over $C$, thus
$\operatorname{acl}_{\omega}\left(Ca\right)$ embeds in $M$ over $C$; using
maximality of $C$ we thus find that $C=B$, and the proof is finished.

######  Remark 9.12.

We now give an alternative approach to the proof of Theorem 9.10, avoiding
transfinite towers. In fact, the argument shows that the transfinite tower
constructed above terminates in finitely many steps, with an explicit bound.

Let $K$ be a countable model of $\operatorname{VFA}$ which is algebraically
closed , transformally Henselian and nontrivially valued. Let $M$ be a
saturated model of $\widetilde{\operatorname{VFA}}$ over $K$, and
$L=K\left(a\right)_{\sigma}$ a transformally algebraic extension of $K$, with
$a$ a single element; we want to embed $L$ in $M$ over $K$. Using Example 5.25
and compactness, this guarantees that every countable transformally algebraic
extension of $K$ embeds in $L$ over $M$.

The element $a$ is a root of some difference polynomial $fx$ with coefficients
in $K$. After twisting, we may assume that $f^{\prime}a\neq 0$. By
scatteredness, only finitely many valuative distances are attained between
distinct roots of $fx$, say $\gamma_{1}<\ldots<\gamma_{N}$. Note that this set
of valuative distances is fixed, and does not change upon base extension. To
facilitate the induction below, we also put formally $\gamma_{0}=-\infty$.

We will find inductively a countable $K_{n}=\operatorname{acl}_{\omega}K_{n}$
over $K$ contained in $M$, an element $a_{n}\in K_{n}$ and common amalgamation
of $L$ and $K_{n}$ over $K$, such that the valuative distance
$v\left(a-a_{n}\right)>\gamma_{n}$ inside the amalgam; i.e, the element $a$
lies in an open ball of valuative radius $\gamma_{n}$ around an element of
$K_{n}$. For $n=0$, we can simply put $K_{0}=K$ and $a_{0}=0$.

Let us assume for that this has been achieved. Relabelling, we may write
$K=K_{n}$. Then $a$ is the unique root of of $fx$ inside some open ball
$K$-ball; since $K$ is transformally Henselian, the element $a$ lies in $K$,
and we are done.

It remains to find $K_{n+1}$ given $K_{n}$. So assume that $K_{n}$ and $c_{n}$
were found; for simplicity of notation, relabelling, we may write $K=K_{n}$,
$c_{n}=c$, $\alpha=\gamma_{n}$ and $\beta=\gamma_{n+1}$. So $a$ lies an open
ball of valuative radius $\alpha$ around an element $K$; we seek to increase
$K$ inside $M$ and amalgamate, so that $a$ will lie in an open ball of
valuative radius $\beta$ around an element of $K$. Since $M$ is transformally
Newtonian, there is an $M$-root $d$ of $fx$ in the open ball of valuative
radius $\alpha$ around the element $a$. Let us amalgamate
$\operatorname{acl}_{\omega}Kc$ and $L$ together over $K$; replacing $K$ by
$\operatorname{acl}_{\omega}Kd$ we may therefore assume that $a$ lies in a
closed ball of valuative radius $\beta$ around an element of $K$. Increasing
$K$ again inside $M$ we may assume that $\Gamma$ is transformally divisible;
after an affine change of variables, we may thus assume that $\beta=0$.

To finish it is enough to find some $d\in\operatorname{acl}_{\omega}Ka$, whose
residue class coincides with that $a$, and such that
$\operatorname{acl}_{\omega}Kd$ embeds in $M$ over $K$. If the residue class
of $a$ has a representative in $K$, then this is trivial. But in any case, the
argument of Lemma 9.11 applies, and the induction step is complete.

######  Proposition 9.13.

Let $K$ be a model of $\widetilde{\operatorname{VFA}}$ and $\kappa$ an
infinite cardinal. Then $K$ is $\kappa^{+}$-saturated as a model of
$\widetilde{\operatorname{VFA}}$ if and only if $k$ and $\Gamma$
$\kappa^{+}$-saturated as models of $\operatorname{ACFA}$ and
$\widetilde{\omega\operatorname{OGA}}$ respectively, and moreover every nested
family of balls in $K$ of cardinality $\kappa$ has a nonempty intersection.

###### Proof.

As in the proof of Theorem 9.10, it is enough to establish the following. Let
$A$ be a small model contained in $K$ and $B$ a small algebraically closed and
transformally Henselian extension of $A$ properly extending $A$; then there is
some $A\subset B^{\prime}\subset B$ with $B^{\prime}$ transformally Henselian,
without any finite $\sigma$-invariant Galois extensions, properly extending
$A$, and so that $B^{\prime}$ embeds in $K$ over $A$.

If $B$ is transformally algebraic over $A$ then $k_{B}$ properly extends
$k_{A}$, as $A$ is transformally Newtonian. By saturation of $k$ we can embed
$k_{B}$ in $k$ over $k_{A}$; this lifts uniquely to an embedding of
$B^{\text{tunr}}$ in $K$ over $A$. The field $B^{\text{tunr}}$ has no
nontrivial finite $\sigma$-invariant Galois extensions, since $k_{B}$ is
algebraically closed. So we may assume that no element of $B$ is transformally
algebraic over $A$ unless it already lies in $A$; thus every element of $B$
not in $A$ is generic over $A$. Saturation of $k$ implies that it is of large
transformal transcendence degree over the prime field, so the generic of
$\mathcal{O}$ over $A$ is realized in $K$. Similarly saturation of $\Gamma$
and the fact that every small nested family of closed balls in $K$ has a
nonempty intersection implies that the other two kinds of generic types over
$A$ are realized in $K$. ∎

######  Theorem 9.14.

Let $K$ be a model of $\widetilde{\operatorname{VFA}}$. Then $k$ and $\Gamma$
are embedded, stably embedded and fully orthogonal pure models of
$\operatorname{ACFA}$ and $\widetilde{\omega\operatorname{OGA}}$ respectively.
In other words, let $D\subset k^{n}\times\Gamma^{m}$ be a definable set which
is definable with parameters in $K$ and in the language of
$\widetilde{\operatorname{VFA}}$. Then $D$ is a boolean combination of
definable sets of the form $X_{i}\times Y_{j}$ where $X_{i}\subset k^{n}$ and
$Y_{j}\subset\Gamma^{m}$ are definable in the language of difference fields
and ordered transformal modules, respectively. Furthermore the $X_{i}$ are
definable with parameters in $k$ and the $Y_{j}$ are definable with parameters
in $\Gamma$.

###### Proof.

We may take $K=k\left(\left(t^{\Gamma}\right)\right)$ where $k,\Gamma$ are
saturated; so by Proposition 9.13 the field $K$ is saturated as a model of
$\widetilde{\operatorname{VFA}}$. The restriction map
$\text{Aut}\left(K\right)\to\text{Aut}\left(k\right)\times\text{Aut}\left(\Gamma\right)$
is evidently surjective, indeed has a section. So $k$ and $\Gamma$ are stably
embedded and fully orthogonal; see e.g [ChaHru99]. We must show that the
induced structure on $k$ and $\Gamma$ is the pure difference field and pure
ordered transformal module structure. By quantifier elimination in
$\operatorname{ACFA}$ and $\widetilde{\omega\operatorname{OGA}}$, it is enough
to prove: if $k_{1}\cong k_{2}$ is an isomorphism between small algebraically
closed difference subfields of $k$ and $\Gamma_{1}\cong\Gamma_{2}$ is an
isomorphism between finitely generated transformal submodules of $\Gamma$ then
it lifts to a global automorphism of $K$. The valuation map and the residue
map both have a section; so we get an isomorphism
$k_{1}\left(t^{\Gamma_{1}}\right)\cong k_{2}\left(t^{\Gamma_{2}}\right)$ of
transformal valued subfields of $K$. We then finish using quantifier
elimination in $\widetilde{\operatorname{VFA}}$, saturation and Corollary
6.17. ∎

######  Proposition 9.15.

The theories
$\widetilde{\operatorname{VFA}},\widetilde{\operatorname{VFA}_{\mathbf{Q}}}$
and $\widetilde{\operatorname{VFA}_{\mathbf{F}_{p}}}$ are recursive.

###### Proof.

Let $\phi$ be a sentence of the language of $\operatorname{VFA}$. If
$\neg\phi$ holds in every model of $\widetilde{\operatorname{VFA}}$ then by
the completeness theorem there is a syntactic proof of $\neg\phi$ derived from
the $\widetilde{\operatorname{VFA}}$ axioms; this is then witnessed computably
as $\widetilde{\operatorname{VFA}}$ is recursively axiomatized. Suppose
conversely that $M$ is a model of $\widetilde{\operatorname{VFA}}$ and
$M\models\phi$. The elementary theory of $M$ as a model of
$\operatorname{VFA}$ is determined by the elementary theory of $k_{M}$ as a
pure difference field and the fact that $M$ is a model of
$\widetilde{\operatorname{VFA}}$. By compactness there is therefore a sentence
$\psi$ consistent with $\operatorname{ACFA}$ such that $\phi$ continues to
hold in every model of $\widetilde{\operatorname{VFA}}$, provided only that
the residue field obeys $\psi$; since $\operatorname{ACFA}$ is recursive,
systematically enumerating all such sentences and searching for a syntactic
proof, we find that if the theory of models of
$\widetilde{\operatorname{VFA}}$ where $\phi$ holds is consistent then this
can be witnessed computably. It follows that $\widetilde{\operatorname{VFA}}$
is recursive; the proof of the other cases goes through without change since
the theories $\operatorname{ACFA}_{\mathbf{Q}}$ and
$\operatorname{ACFA}_{\mathbf{F}_{p}}$ are recursive. ∎

######  Proposition 9.16.

Let $M$ be a model of $\widetilde{\operatorname{VFA}}$ and let $K$ be a
transformal valued subfield. Then $K$ is model theoretically algebraically
closed in $M$ if and only if it is field theoretically algebraically closed,
inversive, and transformally Henselian.

###### Proof.

By uniqueness of Hensel lifts, the transformal Henselization is contained in
the definable closure. The inversive hull and the field theoretic algebraic
closure are moreover evidently contained in the model theoretic algebraic
closure. For the converse, let $E$ be an algebraically closed, transformally
Henselian extension of $K$ in $M$. Then $E\otimes_{K}E$ can be equipped with
the structure of a model of $\operatorname{VFA}$. By model completeness of
$\widetilde{\operatorname{VFA}}$ and quantifier elimination, we find that
$\text{tp}\left(\nicefrac{{E}}{{K}}\right)$ is realized anew over $E$, so that
no element of $E$ is model theoretically algebraic over $K$ unless it already
lies in $K$. ∎

###  9.3. Quantifier Elimination

We give now give a precise version of the quantifier elimination statement.

####  9.3.1.

Let $\operatorname{FA}$ be the theory of perfect, inversive difference fields.
The language $\mathcal{L}_{\operatorname{FA}}$ expands the language of fields
by unary function symbols $\sigma^{\pm 1}$ and $\phi^{\pm 1}$ with the obvious
interpretations. Let $\widetilde{\mathcal{L}_{\operatorname{FA}}}$ be the
expansion of $\mathcal{L}_{\operatorname{FA}}$ where basic formulas are of the
form:

$\exists y\theta\left(x_{1},\ldots,x_{n},y\right)$

where $\theta\left(x_{1},\ldots,x_{n},y\right)$ is a quantifier free formula
in the language $\mathcal{L}_{\operatorname{FA}}$; and moreover for some
$m\in\mathbf{N}$ and difference polynomials
$f_{0},\ldots,f_{m-1},g_{0},\ldots,g_{m-1}\in\mathbf{Z}\left[x_{1},\ldots,x_{n}\right]_{\sigma}$
the formula $\theta\left(x_{1},\ldots,x_{n},y\right)$ implies that:

$y^{m}=f_{0}\left(x_{1},\ldots,x_{n}\right)+\ldots+f_{m-1}\left(x_{1},\ldots,x_{n}\right)\cdot
y^{m-1}$
$y^{\sigma}=g_{0}\left(x_{1},\ldots,x_{n}\right)+g_{m-1}\left(x_{1},\ldots,x_{n}\right)\cdot
y^{m-1}$

This means that quantification is only allowed over algebraic difference field
extensions which are primitive already as abstract field extensions (and in
particular finite).

######  Proposition 9.17.

The theory $\operatorname{ACFA}$ admits elimination of quantifiers in
$\widetilde{\mathcal{L}_{\operatorname{FA}}}$.

###### Proof.

Let $M$ be a saturated model of $\operatorname{ACFA}$ and $\varphi\colon
K\cong L$ an isomorphism between models of $\operatorname{FA}$ contained in
$M$ preserving $\widetilde{\mathcal{L}_{\operatorname{FA}}}$ formulas; we must
prove that $\varphi$ lifts to a global automorphism. Let $\widetilde{K}$ be a
finite Galois extension of $K$ left invariant under $\sigma$ and
$\widetilde{L}$ the corresponding $\sigma$-invariant Galois extension of $L$
under the isomorphism $\varphi$. By the primitive element theorem, we have
$\widetilde{K}=K\left[a\right]=K\left[a\right]_{\sigma}$ for a single element
$a\in\widetilde{K}$. Now $\nicefrac{{\widetilde{K}}}{{K}}$ is finitely
presented as a difference algebra over $K$, so as $\varphi$ preserves
$\widetilde{\mathcal{L}_{\operatorname{FA}}}$ formulas there is some
$b\in\widetilde{L}$ such that $\widetilde{K}$ and $\widetilde{L}$ are
isomorphic as difference fields by an isomorphism extending $\varphi\colon
K\cong L$ and which carries $a$ to $b$. So the isomorphism extends to any
finite $\sigma$-invariant Galois extension.

Let $\overline{K}$ and $\overline{L}$ denote the union of all finite
$\sigma$-invariant Galois extensions of $K$ and $L$ respectively. By the
previous paragraph we can extend $\varphi$ to an isomorphism of any finite
$\sigma$-invariant extension of $K$ contained in $\overline{K}$; by
compactness of the Galois group we can extend $\varphi$ to an isomorphism of
difference fields between $\overline{K}$ and $\overline{L}$. Changing the
notation assume that $K$ has no nontrivial finite $\sigma$-invariant Galois
extensions. In this case there is up to isomorphism a unique field theoretic
algebraic closure of $K$ ∎

####  9.3.2.

Let $\operatorname{VFA}^{h}$ be the theory of algebraically Henselian models
of $\operatorname{VFA}$. The language $\mathcal{L}_{\operatorname{VFA}^{h}}$
can be described as follows. It admits a single sort $K$ together with unary
predicates $K^{\times},\mathcal{O}$ and $\mathcal{M}$ as well as constant
symbols $0,1$. We have binary function symbols $+,-,\cdot\colon K^{2}\to K$
and finally unary function symbols $\sigma^{\pm 1}$ and $\phi^{\pm 1}$.
Finally, for each $n\geq 2$ we have a function
$h_{n-2}\colon\mathcal{M}^{n-2}\to\mathcal{O}$. The axioms assert that $K$ is
a model of $\operatorname{VFA}$ and for all
$t_{2},\ldots,t_{n}\in\mathcal{M}^{n-2}$, if
$x=h_{n-2}\left(t_{2},\ldots,t_{n}\right)$ then we have

$1+x+t_{2}x^{2}+\ldots+t_{n}x^{n}=0$

we call the $h_{n}$ the Hensel lifting functions of $K$. If $K$ is an
algebraically Henselian model of $\operatorname{VFA}$, then $K$ can be
uniquely expanded to a model of $\operatorname{VFA}^{h}$ and conversely.

Now consider the language $\widetilde{\mathcal{L}_{\operatorname{VFA}}}$ where
basic formulas are of the form:

$\phi\left(x_{1},\ldots,x_{n}\right)\wedge\psi\left(x_{1},\ldots,x_{n}\right)$

where $\phi$ is quantifier free in the language of $\operatorname{VFA}^{h}$
and $\psi\left(x_{1},\ldots,x_{n}\right)$ is a formula in the language
$\widetilde{\mathcal{L}_{\operatorname{FA}}}$.

######  Proposition 9.18.

Notation as above. Then $\widetilde{\operatorname{VFA}}$ eliminates
quantifiers in $\widetilde{\mathcal{L}_{\operatorname{VFA}}}$.

###### Proof.

Let $M$ be a saturated model of $\widetilde{\operatorname{VFA}}$. Let $E,F$ be
models of $\operatorname{VFA}^{h}$ which are contained in $M$ and
$\varphi\colon E\cong F$ an isomorphism of transformal valued fields which
preserves $\widetilde{\mathcal{L}_{\operatorname{VFA}}}$-formulas; we must
show that $\varphi$ lifts to a global automorphism. The proof is the same as
Lemma 9.17, using the criterion of 4.29. ∎

###  9.4. The Elementary Theory of the Frobenius Automorphism

Recall that by a Frobenius transformal valued field we mean a transformal
valued field of $K$ of positive characteristic $p>0$ whose distinguished
endomorphism coincides with a power of the Frobenius endomorphism $\phi
x=x^{p}$. We shall regard perfect Frobenius transformal valued fields as
structures for the language $\operatorname{VFA}$ of perfect, inversive
transformal valued fields.

######  Fact 9.19.

Let $\varphi$ be a sentence in the language of difference fields and assume
that $\operatorname{ACFA}\models\varphi$. Then there is an effectively
computable $N=N\left(\varphi\right)$ such that whenever
$\left(K,\sigma\right)$ is an algebraically closed Frobenius difference field
and $x^{\sigma}=x^{q}$ where $q>N$, then
$\left(K,\sigma\right)\models\varphi$.

###### Proof.

This follows from the fact that the constant implicit in the geometric
estimates of [Hru04] is effectively computable. ∎

######  Theorem 9.20.

Let $T$ be the theory of algebraically closed and nontrivially valued
Frobenius transformal valued fields. Then $T$ is decidable, and the theory
$\widetilde{\operatorname{VFA}}$ is the theory of $\omega$-increasing models
of $T$.

###### Proof.

First we show that $\widetilde{\operatorname{VFA}}$ is the theory of
$\omega$-increasing and nontrivially valued models of $T$. Let $K$ be an
$\omega$-increasing and nontrivially valued model of $T$. Then $K$ is a model
of $\widetilde{\operatorname{VFA}}$; it is transformally Henselian by 5.10;
the valuation group is nonzero and transformally divisible by Proposition 2.4;
the residue field is a model of $\operatorname{ACFA}$ by the main theorem of
[Hru04]; and finally, it is easy to check that the additive difference
operators are onto. Conversely, we must show that every model of
$\widetilde{\operatorname{VFA}}$ arises in this way. Now in
$\widetilde{\operatorname{VFA}}$ the residue field is stably embedded and the
induced structure is precisely the pure difference field structure. Moreover,
the completions of $\widetilde{\operatorname{VFA}}$ are determined by the
elementary theory of the residue field as a pure difference field. Since $K$
is elementarily equivalent to its residue field as a difference field, the
result follows from 3.2.

Now we show that $T$ is decidable. Since $\widetilde{\operatorname{VFA}}$ is
decidable, this boils down to the following: given a sentence $\phi$ such that
$\widetilde{\operatorname{VFA}}\models\phi$ find effectively a natural number
$N=N\left(\varphi\right)$ such that $K\models\varphi$, whenever $K$ is an
algebraically closed and nontrivially valued Frobenius transformal valued
field with $x^{\sigma}=x^{q}$ and $q>N$. It is enough to do this for a
sentence $\varphi$ which is part of our axiomatization. For the
$\operatorname{ACFA}$ axioms over the residue field we can use Fact 9.19. Now
consider for instance a part of the axiom scheme asserting that the field is
transformally Henselian. Such an axiom asserts that the Hensel lifting lemma
for difference polynomials supported on a fixed finite subset
$I\subset\mathbf{N}\left[\frac{\sigma}{p^{\infty}}\right]$. If $p=1$, it is
sufficient to choose $q$ large enough so that the substitution $\sigma\mapsto
q$ from $\mathbf{N}\left[\sigma\right]$ to $\mathbf{N}$ is injective when
restricted to $I$; simply take $q$ large enough so it exceeds the coefficients
of all the elements of $I$. In finite characteristic $p$, let $m\in\mathbf{N}$
be the largest power of $p$ which appears in the denominator of the various
$\nu\in I$ one should choose $q$ large so that $\frac{q}{p^{m}}$ is a natural
number, and then apply the same reasoning. ∎

###  9.5. Various Results

######  Proposition 9.21.

Let $K$ be a saturated model of $\widetilde{\operatorname{VFA}}$ and let $F$
be the fixed field of $K$. Then $F$ is a pseudofinite field which inherits
from $K$ the trivial valuation. Furthermore, every definable subset of $F^{n}$
which is definable with parameters in $K$ and in the language of
$\operatorname{VFA}$ is already definable with parameters in $F$ and in the
pure field language.

###### Proof.

By Hensel lifting the residue map identifies $F$ with the fixed field of $k$,
so the Theorem follows from stable embededness of the residue field in
$\widetilde{\operatorname{VFA}}$ and the corresponding statement for
$\operatorname{ACFA}$; see [ChaHru99]. ∎

######  Proposition 9.22.

Let $K$ be a model of $\widetilde{\operatorname{VFA}}$ and $k_{0}$ the
algebraic closure of the prime field.

(1) Let $0<n\in\mathbf{N}$ and $m\in\mathbf{Z}$; then
$\left(K,\sigma^{n}\circ\phi^{m}\right)$ is a model of
$\widetilde{\operatorname{VFA}}$, which is saturated provided that
$\left(K,\sigma\right)$ is.

(2) Let us assume that $K$ is saturated. Fix $n>0$ and let $\tau_{0}$ be an
automorphism of $k_{0}$ such that $\tau_{0}^{n}$ and $\sigma$ agree on
$k_{0}$; then there is an $\omega$-increasing automorphism $\tau$ of $K$
extending $\tau_{0}$, which renders $\left(K,\tau\right)$ a saturated model of
$\widetilde{\operatorname{VFA}}$, and such that $\tau^{n}=\sigma$ on $K$.

(3) The elementary theory of $K$ is determined by the isomorphism type of the
restriction of $\sigma$ to $k_{0}$.

###### Proof.

For (3) use Theorem 9.10. For (1), let $\widetilde{K}$ denote the model of
$\operatorname{VFA}$ whose underlying valued field is the same and whose
automorphism is given by $\tau$. Then $\widetilde{k}$ is a model of
$\operatorname{ACFA}$ which is saturated provided that $k$ is; see [ChaHru99],
1.12. Moreover $\widetilde{\Gamma}\models\widetilde{\omega\operatorname{OGA}}$
provided that $\Gamma\models\widetilde{\omega\operatorname{OGA}}$; by
$o$-minimality these two models are simultaneously saturated. Finally the
field $\widetilde{K}$ is transformally Henselian and the additive difference
operators are onto; so $\widetilde{K}$ is a model of
$\widetilde{\operatorname{VFA}}$. For the second statement use [ChaHru99]
1.12, stable embededness of the residue field and the saturation criterion of
Proposition 9.13. ∎

######  Proposition 9.23.

Let $K$ be a model of $\operatorname{VFA}$ which is nontrivially valued. Then
$K$ is a model of $\widetilde{\operatorname{VFA}}$ if and only if is
transformally Henselian and its underlying difference field is a model of
$\operatorname{ACFA}$.

###### Proof.

If $K$ is a model of $\widetilde{\operatorname{VFA}}$ then $K$ is elementarily
equivalent to its residue field as a difference field by Theorem 9.20, so the
underlying difference field of a model of $\widetilde{\operatorname{VFA}}$ is
a model of $\operatorname{ACFA}$. For the converse, assume
$K\models\operatorname{ACFA}$ is nontrivially valued and transformally
Henselian. Since $K$ is transformally Henselian, the residue field lifts, and
we can identify $k$ with a relatively transformally algebraically closed
difference subfield of $K$; a difference field transformally algebraically
closed inside a model of $\operatorname{ACFA}$ is a model of
$\operatorname{ACFA}$, so $K$ is strictly transformally Henselian. Moreover,
every nonconstant difference polynomial has a root; so
$\left(K^{\times}\right)^{\nu}=K^{\times}$ for all
$0\neq\nu\in\mathbf{Z}\left[\sigma\right]$. Finally, all additive difference
operators are onto, thus $K\models\widetilde{\operatorname{VFA}}$. ∎

######  Proposition 9.24.

Let $\left(K_{20},K_{21},K_{10}\right)$ be a triplet of models of
$\operatorname{VFA}$ with $K_{21}$ and $K_{10}$ nontrivially valued. Then
$K_{20}$ is a model of $\widetilde{\operatorname{VFA}}$ if and only if
$K_{21}$ and $K_{10}$ are so. In particular, the transformally Archimedean
ultrapower of a transformally Archimedean model of
$\widetilde{\operatorname{VFA}}$ is a model of
$\widetilde{\operatorname{VFA}}$.

###### Proof.

Let us assume that $K_{20}\models\widetilde{\operatorname{VFA}}$. Then the
abstract difference field $K_{2}\models\operatorname{ACFA}$ and $K_{21}$ is
transformally Henselian by Lemma 5.12; as $K_{21}$ is by assumption
nontrivially valued, it is a model of $\widetilde{\operatorname{VFA}}$ using
Proposition 9.23. Moreover $K_{1}$ is isomorphic to a difference subfield of
$K_{2}$ which is transformally algebraically closed in $K_{2}$; so $K_{1}$ is
a model of $\operatorname{ACFA}$. Since $K_{10}$ is transformally Henselian,
it follows that $K_{1}$ is a model of $\operatorname{ACFA}$. For the converse,
we know that $K_{20}$ is transformally Henselian so it is enough to prove that
$K_{2}\models\operatorname{ACFA}$; but this follows from the fact that
$K_{21}\models\widetilde{\operatorname{VFA}}$. ∎

######  Proposition 9.25.

Let $K$ be a model of $\widetilde{\operatorname{VFA}}$. Then the completion
$\widehat{K}$ of $K$ is a model of $\widetilde{\operatorname{VFA}}$.

###### Proof.

By Proposition 5.13 the field $\widehat{K}$ is transformally Henselian; the
residue field and the valuation groups of $K$ and $\widehat{K}$ coincide so it
is enough to prove that the additive difference operators are onto on
$\widehat{K}$. Since $\widehat{K}$ is perfect and inversive, after twisting,
we may assume that $\tau^{\prime}\neq 0$. By scatteredness and rescaling we
may assume that all the roots of $\tau x$ are integral. Since $\mathcal{O}$
lies dense in $\widehat{\mathcal{O}}$, approximating the coefficients we can
find $a\in\mathcal{O}$ with $v\tau a>\beta$ where $\beta$ is the valuation of
the coefficient of the constant term; since $\widehat{K}$ is transformally
Henselian, by Lemma 5.11 we find that $\tau$ admits a root in
$\widehat{\mathcal{O}}$. ∎

######  Corollary 9.26.

Let $k$ be a model of $\operatorname{ACFA}$ and $t$ an element transformally
transcendental over $k$. Then there is a model $K\models\operatorname{ACFA}$
over $k\left(t\right)_{\sigma}$ and such that no element of $K$ is
transformally algebraic over $k$, unless it already lies in $k$.

###### Proof.

Let us regard $k$ as a trivially valued model of $\operatorname{VFA}$ and set
$K=k\left(\left(t^{\mathbf{Q}\left(\sigma\right)}\right)\right)$; then
$K\models\widetilde{\operatorname{VFA}}$ and hence
$K\models\operatorname{ACFA}$ by Proposition 9.23. ∎

######  Example 9.27.

If $K$ is an algebraically closed and nontrivially valued field, then $K$ is
existentially closed as a valued field and in particular Henselian. This is no
longer true in the presence of an automorphism; the requirement that $K$ be
transformally Henselian in Proposition 9.23 is necessary. For example, let
$K\models\widetilde{\operatorname{VFA}}$. Let $x=a+t$ where $a^{\sigma}=a$ and
$t$ is generic in $\mathcal{M}$ over $K$; assume moreover the residue class of
$a$ is algebraically transcendental over $k$. Let $L$ be a model of
$\operatorname{ACFA}$ over $K\left(x\right)_{\sigma}$ with no element of $L$
transformally algebraic over $K$, unless it already lies in $K$. The fixed
field of $L$ must then coincide with the fixed field of $K$. On the other
hand, the fixed field of $l$ properly extends the fixed field of $k$. So the
fixed field of $L$ does not lift to a trivially valued field; it follows that
$L$ cannot be transformally Henselian.

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