# Decomposition of first order Lipschitz functions by Clifford algebra-valued
harmonic functions

Lianet De la Cruz Toranzo  Department of Mathematics and Informatics, TU
Bergakademie Freiberg, Germany.  Ricardo Abreu Blaya  Facultad de Matemáticas,
Universidad Autónoma de Guerrero, México.  Investigador Invitado. Universidad
UTE, Ecuador.  Swanhild Bernstein  Department of Mathematics and Informatics,
TU Bergakademie Freiberg, Germany.

###### Abstract

In this paper we solve the problem on finding a sectionally Clifford algebra-
valued harmonic function, zero at infinity and satisfying certain boundary
value condition related to higher order Lipschitz functions. Our main tool are
the Hardy projections related to a singular integral operator arising in
bimonogenic function theory, which turns out to be an involution operator on
the first order Lipschitz classes. Our result generalizes the classical Hardy
decomposition of Hölder continuous functions on a simple closed curve in the
complex plane.

##  1 Introduction

A classical boundary value problem in Complex analysis is to find a
sectionally holomorphic function $\Phi(z)$, zero at infinity and satisfying
the given boundary condition $\Phi(t)^{+}-\Phi(t)^{-}=\varphi(t),$
$\,t\in\Gamma,$ where $\varphi$ satisfies the Hölder condition on $\Gamma$,
i.e. $\varphi\in{\bf C}^{0,\alpha}(\Gamma)=\\{f:|f(t)-f(\tau)|\leq
c|t-\tau|^{\alpha};\,t,\tau\in\Gamma\\},\,0<\alpha\leq 1.$ Here $\Gamma$ is a
closed smooth curve in the complex plane and $\Phi(t)^{\pm}$ are the limiting
values as approaching the curve from the interior $\Omega_{+}$ and the
exterior $\Omega_{-}$ domain respectively. This problem is immediately solved
by the Cauchy transform,

$\mathcal{C}\varphi(z):=\frac{1}{2\pi
i}\int_{\Gamma}\frac{\varphi(\zeta)}{\zeta-z}d\zeta,z\in\mathbb{C}\setminus\Gamma,$

which has continuous limiting values as approaching $\Gamma$ from the interior
or exterior domain [3, 5]. Those involve both the identity and the singular
integral operator

$\mathcal{S}\varphi(t)=p.v.\frac{1}{\pi
i}\\!\int_{\Gamma}\\!\frac{\varphi(\zeta)}{\zeta-t}d\zeta=\lim\limits_{\epsilon\to
0}\frac{1}{\pi
i}\\!\int_{\Gamma\setminus\Gamma_{\epsilon}}\\!\\!\frac{\varphi(\zeta)}{\zeta-t}d\zeta,\,\,t\in\Gamma,\,\,\Gamma_{\epsilon}=B_{\epsilon}(t)\cap\Gamma$

and they are expressed by the classical Plemelj-Sokhotski formula:

$\begin{cases}\displaystyle\mathcal{C}^{+}\varphi(t)=\lim\limits_{z\to t\atop
z\in\Omega_{+}}\mathcal{C}\varphi(z)=\frac{1}{2}\varphi(t)+\frac{1}{2\pi
i}\int_{\Gamma}\frac{\varphi(\zeta)}{\zeta-t}d\zeta=\frac{1}{2}\big{[}I+\mathcal{S}\big{]}\varphi,\\\
\displaystyle\mathcal{C}^{-}\varphi(t)=\lim\limits_{z\to t\atop
z\in\Omega_{-}}\mathcal{C}\varphi(z)=-\frac{1}{2}\varphi(t)+\frac{1}{2\pi
i}\int_{\Gamma}\frac{\varphi(\zeta)}{\zeta-t}d\zeta=\frac{1}{2}\big{[}-I+\mathcal{S}\big{]}\varphi.\end{cases}$
(1)

The limiting values of the Cauchy transform with density function satisfying
the Hölder condition, also verify this [3, 5]. Therefore, the singular
integral operator keeps invariant the Hölder class (Plemelj-Privalov theorem)
and it yields the Hardy decomposition of Hölder functions: ${\bf
C}^{0,\alpha}(\Gamma)={\bf C}^{0,\alpha}(\Gamma)^{+}\oplus{\bf
C}^{0,\alpha}(\Gamma)^{-},$ where the uniqueness may be seen as a consequence
of the involution property for the singular operator, i.e.
$\mathcal{S}^{2}=I$.

In the present work, we aim to obtain a similar decomposition for first order
Lipschitz functions in the framework of Clifford analysis instead, which is a
multidimensional function theory where the so-called monogenic functions play
a similar roll to that by the holomorphic ones in Complex analysis. Our main
result is Theorem 2 and it will be stated and proved in Section 4.

##  2 Preliminaries

###  2.1 Clifford algebras and Clifford analysis

Denote by $\\{e_{i}\\}_{i=1}^{m}$ an orthonormal basis of $\mathbb{R}^{m}$
governed by the multiplication rules

$\displaystyle e_{i}^{2}=-1,\quad e_{i}e_{j}=-e_{j}e_{i},\,\,\forall
i,j=1,2,\dots,m;\,\,i<j.$ (2)

Then the real Clifford algebra $\mathbb{R}_{0,m}$ is generated by
$\\{e_{i}\\}_{i=1}^{m}$ over the field of real numbers $\mathbb{R}$, so that
an element $a\in\mathbb{R}_{0,m}$ may be written as

$a=\sum_{A}a_{A}e_{A},$

where $a_{A}\in\mathbb{R}$ and $A$ runs over all the possible ordered sets
$A=\\{1\leq i_{1}<\cdots<i_{k}\leq m\\}$ or $A=\emptyset$ and
$e_{A}:=e_{i_{1}}e_{i_{2}}\cdots e_{i_{k}},\,\,e_{0}=e_{\emptyset}=1.$ If we
identify $\underline{x}\in\mathbb{R}^{m}$ by
$\underline{x}=x_{1}e_{1}+\dots+x_{m}e_{m};\,x_{i}\in\mathbb{R},i=\overline{1,m}$
then we easily see that $\mathbb{R}^{m}$ is embedded in $\mathbb{R}_{0,m}$.

A norm for $a\in\mathbb{R}_{0,m}$ may be defined by
$\|a\|^{2}=\sum_{A}|a_{A}|^{2},$ so that the Clifford algebra can be seen as a
$2^{n}$-Euclidean space with Euclidean metric. In particular, for
$\underline{x}\in\mathbb{R}^{m}$ we have $\|\underline{x}\|=|\underline{x}|$,
where $|\cdot|$ denotes the usual Euclidean norm.

We will consider functions defined on subsets of $\mathbb{R}^{m}$ and taking
values in $\mathbb{R}_{0,m}$, namely $f=\sum_{A}f_{A}e_{A}$. We shall say that
an $\mathbb{R}_{0,m}$-valued function belongs to certain classical class of
functions, if each of its real components $f_{A}$ do so.

From now on, $\Omega$ stands for a Jordan domain, i.e. a bounded oriented
connected open subset of $\mathbb{R}^{m}$ whose boundary is a compact
topological surface. For simplicity, we assume $\partial\Omega$ to be
sufficiently smooth, e.g. Liapunov surface. By $\Gamma$ we denote the boundary
of $\Omega$, while $\Omega_{+}=\Omega$,
$\Omega_{-}=\mathbb{R}^{m}\setminus{\Omega\cup\Gamma}$ if necessary.

An $\mathbb{R}_{0,m}$-valued function $f$ in ${\bf C}(\Omega)$ is called left
(resp. right) monogenic if ${\cal D}_{\underline{x}}\,f=0$ (resp. $f\,{\cal
D}_{\underline{x}}=0$) in $\Omega$, where ${\cal D}_{\underline{x}}$ is the
Dirac operator

${\cal
D}_{\underline{x}}={\partial_{x_{1}}}e_{1}+{\partial_{x_{2}}}e_{2}+\cdots+{\partial_{x_{m}}}e_{m}.$

It is easy to check that the so-called Clifford-Cauchy kernel
$E_{0}(\underline{x})=-\displaystyle\frac{1}{\sigma_{m}}\frac{\underline{x}}{|\underline{x}|^{m}}\,\,(\underline{x}\neq
0),$ where $\sigma_{m}$ stands for the surface area of the unit sphere in
$\mathbb{R}^{m}$, is a two-sided monogenic function.

If $f\in{\bf C}^{1}(\Omega_{+}\cup\Gamma)$ is monogenic in $\Omega_{+}$, then
the following representation formula holds

$f(\underline{x})=\int_{\Gamma}E_{0}(\underline{y}-\underline{x})n(\underline{y})f(\underline{y})d\underline{y},\quad\underline{x}\in\Omega_{+},$

where $n(\underline{y})$ stands for the outward pointing unit normal at
$\underline{y}\in\Gamma.$

This formula gives rise to the cliffordian Cauchy type transform and its
singular version

$\mathcal{C}_{0}=\int_{\Gamma}E_{0}(\underline{y}-\underline{x})n(\underline{y})f(\underline{y})d\underline{y},\,(\underline{x}\in\Omega_{+})\quad\mbox{and}\quad\mathcal{S}_{0}=2\int_{\Gamma}E_{0}(\underline{y}-\underline{x})n(\underline{y})f(\underline{y})d\underline{y},\,(\underline{x}\in\Gamma).$

When $f$ satisfies the Hölder condition then, as in (1), both operators are
connected by the cliffordian Plemelj-Sokhotski formula [4], namely

$[\mathcal{C}_{0}f]^{+}=\frac{1}{2}[I+\mathcal{S}_{0}]f\quad\mbox{and
}\quad[\mathcal{C}_{0}f]^{-}=\frac{1}{2}[-I+\mathcal{S}_{0}]f.$ (3)

Note that ${\cal D}_{\underline{x}}$ factorizes the Laplace operator
$\triangle_{m}$ in $\mathbb{R}^{m}$ in the sense that ${\cal
D}_{\underline{x}}^{2}=-\triangle_{m}.$ The fundamental solution of ${\cal
D}_{\underline{x}}$ is thus given by $E_{0}(\underline{x})={\cal
D}_{\underline{x}}E_{1}(\underline{x}),$ where

$E_{1}(\underline{x})=\frac{1}{(m-2)\sigma_{m}{|\underline{x}|}^{m-2}}\,\,(\underline{x}\neq
0,\,m>2)$

is the fundamental solution of the Laplacian in $\mathbb{R}^{m}$ ($m\geq 3$).

In addition, an $\mathbb{R}_{0,m}$-valued function $f$ in ${\bf
C}^{k}(\Omega)$ is called polymonogenic of order $k$ or simply $k$-monogenic
(left) if it satisfies ${\cal D}_{\underline{x}}^{k}\,f=0$ in $\Omega$.

###  2.2 Higher order Lipschitz functions and Whitney extension theorem

Given an appropriate space of functions defined on a non-empty closed set
${\bf E}\subset\mathbb{R}^{m}$, how can these be extended to $\mathbb{R}^{m}$?
The study of these problems and linear properties of some related extension
operators in terms of Banach spaces, led to the higher order Lipschitz
functions, defined by E. Stein in [7], but going back to the works of H.
Whitney [8], who introduced a notion of differentiation in a general set
(${\bf C}^{m}$-continuity in terms of $f_{(k)}$). The most appropriate spaces
are the ones given in terms of the modulus of continuity, in particular, the
Lipschitz spaces ${\mbox{Lip}}(\alpha,{\bf E})$ with exponent $0<\alpha\leq 1$
composed by those functions $f$ defined on ${\bf E}$ such that

$\displaystyle|f(x)|\leq M\quad\mbox{and}\quad|f(x)-f(y)|\leq
M|x-y|^{\alpha}\quad\mbox{for}\quad x,\,y\in{\bf E},$ (4)

which are Banach spaces with the smallest $M$ in the above definition as norm.
In fact, the linear extension operator ${\cal{E}}_{0}$ defined by

${\cal{E}}_{0}(f)(x)=\begin{cases}f(x),&x\in{\bf E}\\\
\sum\limits_{i}f(p_{i})\varphi_{i}^{*}(x),&x\in{\bf E}^{c}\end{cases}$ (5)

maps ${\mbox{Lip}}(\alpha,{\bf E})$ continuously into
${\mbox{Lip}}(\alpha,\mathbb{R}^{m})$ for $0<\alpha\leq 1$ and the norm is
independent of the closed set ${\bf E}$. This definition deals with the ideas
on decomposition of open sets into cubes and partition of the unity as well.
It calls for more explanation, but it is beyond the scope of this paper. For
more details we refer the reader to [7, Ch. VI].

When $\alpha>1$, the space ${\mbox{Lip}}(\alpha,{\bf E})$ is constituted by
constants only. However, it can be generalized as follows.

######  Definition 1 (Higher order Lipschitz functions).

Let ${\bf E}$ be a closed subset of $\mathbb{R}^{m}$, $k$ a nonnegative
integer and $0<\alpha\leq 1$. We shall say that a real valued function $f$,
defined in ${\bf E}$, belongs to the higher order Lipschitz class
${\mbox{Lip}}(k+\alpha,{\bf E})$ if there exist real valued functions
$f^{(j)}$, $0<|j|\leq k$, defined on ${\bf E}$, with $f^{(0)}=f$, and so that
if

$f^{(j)}(x)=\sum_{|j+l|\leq
k}\frac{f^{(j+l)}(y)}{l!}(x-y)^{l}+R_{j}(x,y),\,\,x,y\in{\bf E}$

then

$|f^{(j)}(x)|\leq M,\,\,\,|R_{j}(x,y)|\leq
M|x-y|^{k+\alpha-|j|},\,\,x,y\in{\bf E},|j|\leq k,$

where $M$ is a positive constant and $j=(j_{1},\dots,j_{m})$ and
$l=(l_{1},\dots,l_{m})$ are $m$-dimensional multi-indices in
${\mathbb{N}}_{0}^{m}$.

We recall the multi-indices conventions

$\underline{x}^{j}:=x_{1}^{j_{1}}\cdots x_{m}^{j_{m}},\,j!:=j_{1}!\cdots
j_{m}!,\,|j|:=j_{1}+\cdots+j_{m},\,\partial^{j}:=\frac{\partial^{|j|}}{\partial^{j_{1}}_{x_{1}}\dots\partial^{j_{m}}_{x_{m}}}.$

First, some remarks concerning this definition are in order. We note that the
function $f^{(0)}=f$ does not necessarily determine the elements $f^{(j)}$ for
an arbitrary ${\bf E}$. Therefore, an element of ${\mbox{Lip}}(k+\alpha,{\bf
E})$ has to be interpreted as a collection of functions $\\{f^{(j)}:{\bf
E}\mapsto\mathbb{R},\,|j|\leq k\\}$. However, if ${\bf E}=\mathbb{R}^{m}$ and
$f\in{\mbox{Lip}}(k+\alpha,\mathbb{R}^{m})$, then according to the definition
$f$ is continuous and bounded and has continuous and bounded partial
derivatives $\partial^{j}f$ up to the order $k$ and the functions
$f^{(j)}:=\partial^{j}f$ for $|j|=k$ belong to the space
${\mbox{Lip}}(\alpha,\mathbb{R}^{m})$. Moreover, when $k=0$, the Lipschitz
class becomes the usual class ${\bf C}_{b}^{0,\alpha}({\bf E})$ of bounded
Hölder continuous functions in ${\bf E}$, which is consistent with (4).

A generalization of (5) is given by the linear extension operator

${\cal{E}}_{k}(f^{(j)})(x)=\begin{cases}f^{(0)}(x),&x\in{\bf E}\\\
{\sum\limits_{i}}^{\prime}P(x,p_{i})\varphi_{i}^{*}(x),&x\in{\bf
E}^{c}\end{cases}$ (6)

where $P(x,y)$ denotes the Taylor expansion of $f$ about $y\in{\bf E},$ i.e.
$P(x,y)=\displaystyle\sum_{|l|\leq k}\frac{f^{(l)}(y)}{l!}(x-y)^{l}$.

For $0<\alpha\leq 1$, the operator ${\cal E}_{k}$ maps
${\mbox{Lip}}(k+\alpha,{\bf E})$ continuously into
${\mbox{Lip}}(k+\alpha,\mathbb{R}^{m})$ and the norm is independent of the
closed set ${\bf E}$.

By using the Clifford norm $\|\cdot\|$ the $\mathbb{R}_{0,m}$-valued class of
functions ${\mbox{Lip}}(k+\alpha,{\bf E})$ may also be defined by means of the
corresponding compatibility conditions

$R_{j}(\underline{x},\underline{y})=f^{(j)}(\underline{x})-\sum_{|l|\leq
k-|j|}\frac{f^{(j+l)}(\underline{y})}{l!}(\underline{x}-\underline{y})^{l},\,\,\underline{x},\underline{y}\in{\bf
E}$ $\|f^{(j)}(\underline{x})\|\leq
M,\;\|R_{j}(\underline{x},\underline{y})\|\leq
M|\underline{x}-\underline{y}|^{k+\alpha-|j|},\,\,\underline{x},\underline{y}\in{\bf
E},|j|\leq k,$

where the functions $f^{(j)}$ and $R_{j}(\underline{x},\underline{y})$ are
$\mathbb{R}_{0,m}$-valued as well.

As we mentioned before, the higher order Lipschitz class is connected with the
works of H. Whitney, namely his celebrated extension theorem [8, Thm. I],
which turns out to be an appropriate way to define our multidimensional
singular integral operator later.

######  Theorem 1 (Whitney Extension Theorem).

Let $f\in{\mbox{Lip}}(k+\alpha,{\bf E})$. Then, there exists a function
$\tilde{f}\in{\mbox{Lip}}(k+\alpha,\mathbb{R}^{m})$ satisfying

  * (i)

$\tilde{f}|_{\bf E}=f^{(0)}$,

  * (ii)

$\partial^{j}\tilde{f}|_{\bf E}=f^{(j)}$, $0<|j|\leq k$,

  * (iii)

$\tilde{f}\in{\bf C}^{\infty}(\mathbb{R}^{m}\setminus{\bf E})$.

##  3 Auxiliary results

Our focus in this paper will be on the particular case of bimonogenic
functions (i.e. $k$-monogenic with $k=2$), which are nothing more than
$\mathbb{R}_{0,m}$-valued harmonic functions.

Bimonogenic functions enjoy a representation formula [6, Theorem 7], namely if
$f$ is bimonogenic in $\Omega$, then

$f(\underline{x})=\int_{\Gamma}E_{0}(\underline{y}-\underline{x})n(\underline{y})f(\underline{y})d\underline{y}-\int_{\Gamma}E_{1}(\underline{y}-\underline{x})n(\underline{y}){\cal
D}_{\underline{y}}f(\underline{y})d\underline{y},x\in\Omega.$ (7)

Now assume $f\in{\bf C}^{2}(\Omega_{-})\cap{\bf C}^{1}(\Omega_{-}\cup\Gamma)$
is bimonogenic in $\Omega_{-}$, $f(\infty)$ exists and ${\cal
D}_{\underline{y}}f(\underline{y})=\scriptscriptstyle\mathcal{O}\big{(}\displaystyle\frac{1}{|\underline{y}|}\big{)}$
as $|\underline{y}|\to\infty$. Consider the ball $B_{R}(\underline{x})$ with
center in $\underline{x}\in\Omega_{-}$ and radius $R$ sufficiently big such
that $\Omega_{+}\cup\Gamma\subset B_{R}(\underline{x})$, then if (7) is
applied to the domain $B_{R}(\underline{x})\setminus\Omega_{+}\cup\Gamma$ we
obtain

$f(\underline{x})=\int_{\Gamma^{*}}E_{0}(\underline{y}-\underline{x})n(\underline{y})f(\underline{y})d\underline{y}-\int_{\Gamma^{*}}E_{1}(\underline{y}-\underline{x})n(\underline{y}){\cal
D}_{\underline{y}}f(\underline{y})d\underline{y},$

where $\Gamma^{*}=-\Gamma\cup C_{R}(\underline{x})$ and
$C_{R}(\underline{x})=\partial B_{R}(\underline{x})$. Then it follows from the
identity

$\int_{C_{R}(\underline{x})}E_{0}(\underline{y}-\underline{x})n(\underline{y})f(\underline{y})d\underline{y}=\int_{C_{R}(\underline{x})}E_{0}(\underline{y}-\underline{x})n(\underline{y})[f(\underline{y})-f(\infty)]d\underline{y}+\Big{(}\int_{C_{R}(\underline{x})}E_{0}(\underline{y}-\underline{x})n(\underline{y})d\underline{y}\Big{)}f(\infty),$

the continuity of $f$ and the fact that
$\int_{C_{R}(\underline{x})}E_{0}(\underline{y}-\underline{x})n(\underline{y})d\underline{y}=1$
that

$\int_{C_{R}(\underline{x})}E_{0}(\underline{y}-\underline{x})n(\underline{y})f(\underline{y})d\underline{y}\to
f(\infty),\,\,\mbox{as}\,\,R\to\infty.$

On the other hand, our assumption on the behavior at infinity implies
$|\underline{y}|\|{\cal D}_{\underline{y}}f(\underline{y})\|\leq\epsilon$ as
$|\underline{y}|\to\infty$ and it does so $\|{\cal
D}_{\underline{y}}f(\underline{y})\|\to 0$ as $|\underline{y}|\to\infty$,
hence $|\underline{y}-\underline{x}|\|{\cal
D}_{\underline{y}}f(\underline{y})\|\leq|\underline{y}|\|{\cal
D}_{\underline{y}}f(\underline{y})\|+|\underline{x}|\|{\cal
D}_{\underline{y}}f(\underline{y})\|\to 0$ as $|\underline{y}|\to\infty$, that
is, ${\cal
D}_{\underline{y}}f(\underline{y})=\scriptscriptstyle\mathcal{O}\big{(}\frac{1}{|\underline{y}-\underline{x}|}\big{)}$
as $|\underline{y}|\to\infty$ and therefore

$\Big{\|}\int_{C_{R}(\underline{x})}E_{1}(\underline{y}-\underline{x})n(\underline{y}){\cal
D}_{\underline{y}}f(\underline{y})d\underline{y}\Big{\|}\leq\int_{C_{R}(\underline{x})}|E_{1}(\underline{y}-\underline{x})|\|{\cal
D}_{\underline{y}}f(\underline{y})\|d\underline{y}\leq\frac{c\epsilon}{R^{m-1}}\int_{C_{R}(\underline{x})}d\underline{y}\to
0\,(\epsilon\to 0,\,R\to\infty).$

Under the above-mentioned assumptions, we arrive at a representation formula
in the exterior domain

$\displaystyle
f(\underline{x})=-\int_{\Gamma}E_{0}(\underline{y}-\underline{x})n(\underline{y})f(\underline{y})d\underline{y}+\int_{\Gamma}E_{1}(\underline{y}-\underline{x})n(\underline{y}){\cal
D}_{\underline{y}}f(\underline{y})d\underline{y}+f(\infty),\,\,\underline{x}\in\Omega_{-}.$
(8)

Combining (7) with the Whitney extension theorem (component-wise applied to
$\mathbb{R}_{0,m}$-valued functions) we introduce a Cauchy type transform of a
first order Lipschitz function by

$[\mathcal{C}_{1}f]^{(0)}(\underline{x})=\int_{\Gamma}E_{0}(\underline{y}-\underline{x})n(\underline{y})\tilde{f}(\underline{y})d\underline{y}-\int_{\Gamma}E_{1}(\underline{y}-\underline{x})n(\underline{y}){\cal
D}_{\underline{y}}\tilde{f}(\underline{y})d\underline{y},\,\,\underline{x}\in\mathbb{R}^{m}\setminus\Gamma$

and we do so its singular version

$[\mathcal{S}_{1}f]^{(0)}(\underline{z})=2\int_{\Gamma}E_{0}(\underline{y}-\underline{z})n(\underline{y})\tilde{f}(\underline{y})d\underline{y}-\int_{\Gamma}E_{1}(\underline{y}-\underline{z})n(\underline{y}){\cal
D}_{\underline{y}}\tilde{f}(\underline{y})d\underline{y},\,\,\underline{z}\in\Gamma$
(9)

where $\tilde{f}$ denotes the $\mathbb{R}_{0,m}$-valued Whitney extension of
$f\in{\mbox{Lip}}(1+\alpha,\Gamma)$. Note that $[\mathcal{C}_{1}f]^{(0)}$ is
bimonogenic in $\mathbb{R}^{m}\setminus\Gamma$.

Due to the weakly singularity of the kernel $E_{1}$ we have from (3)

$\displaystyle\begin{cases}\big{(}[\mathcal{C}_{1}f]^{(0)}\big{)}^{+}(\underline{z})-\big{(}[\mathcal{C}_{1}f]^{(0)}\big{)}^{-}(\underline{z})=\tilde{f}|_{\Gamma}=f^{(0)}(\underline{z}),&\,\underline{z}\in\Gamma\\\
\big{(}[\mathcal{C}_{1}f]^{(0)}\big{)}^{+}(\underline{z})+\big{(}[\mathcal{C}_{1}f]^{(0)}\big{)}^{-}(\underline{z})=[\mathcal{S}_{1}f]^{(0)}(\underline{z}),&\,\underline{z}\in\Gamma\end{cases}$
(10)

where, as usual,
$\big{(}[\mathcal{C}_{1}f]^{(0)}\big{)}^{\pm}(\underline{z})=\lim\limits_{\underline{x}\to\underline{z}\atop\underline{x}\in\Omega_{\pm}}[\mathcal{C}_{1}f]^{(0)}(\underline{x}).$

If we set

$\displaystyle[\mathcal{S}_{1}f]^{(j)}(\underline{z})=2\int_{\Gamma}E_{0}^{(j)}(\underline{y}-\underline{z})n(\underline{y})R(\underline{y},\underline{z})d\underline{y}-2\int_{\Gamma}E_{1}^{(j)}(\underline{y}-\underline{z})n(\underline{y}){\cal
D}_{\underline{y}}R(\underline{y},\underline{z})d\underline{y}+f^{(j)}(\underline{z}),\,|j|=1,$

then it follows that the first order Lipschitz class
${\mbox{Lip}}(1+\alpha,\Gamma)$ behaves invariant under the action of this
singular integral operator [2], that is, the Whitney data
$\mathcal{S}_{1}f:=\\{[\mathcal{S}_{1}f]^{(j)},\,0\leq|j|\leq 1\\}$ belongs to
${\mbox{Lip}}(1+\alpha,\Gamma)$, whenever $f\in{\mbox{Lip}}(1+\alpha,\Gamma)$.

We include for later reference the following lemma which was recently proved
in [1].

######  Lemma 1.

Let $f,g\in{\mbox{Lip}}(1+\alpha,\Gamma)$ such that
$f^{(0)}(\underline{x})=g^{(0)}(\underline{x})$ and
$\sum\limits_{|j|=1}e_{(j)}f^{(j)}(\underline{x})=\sum\limits_{|j|=1}e_{(j)}g^{(j)}(\underline{x})$
for all $\underline{x}\in\Gamma$. Then $f\equiv g$ in $\Gamma$, i.e.
$f^{(j)}=g^{(j)}\,\,\forall|j|\leq k.$

##  4 Main results

######  Theorem 2.

The singular integral operator $\mathcal{S}_{1}$ is an involution. Namely,

$[\mathcal{S}_{1}^{2}f]^{(j)}=f^{(j)}\,\,\forall\,\,|j|\leq 1.$ (11)

###### Proof.

Let us first show that

$\displaystyle\sum\limits_{|j|=1}e_{(j)}[\mathcal{S}_{1}f]^{(j)}(\underline{z})=2\int_{\Gamma}E_{0}(\underline{y}-\underline{z})n(\underline{y})\Big{(}\sum\limits_{|j|=1}e_{(j)}f^{(j)}(\underline{y})\Big{)}d\underline{y}.$
(12)

Indeed,

$\displaystyle\sum\limits_{|j|=1}e_{(j)}[\mathcal{S}_{1}f]^{(j)}(\underline{z})$
$\displaystyle=$ $\displaystyle
2\int_{\Gamma}\sum\limits_{|j|=1}e_{(j)}E_{0}^{(j)}(\underline{y}-\underline{z})n(\underline{y})R(\underline{y},\underline{z})d\underline{y}-2\int_{\Gamma}\sum\limits_{|j|=1}e_{(j)}E_{1}^{(j)}(\underline{y}-\underline{z})n(\underline{y}){\cal
D}_{\underline{y}}R(\underline{y},\underline{z})d\underline{y}$ (13)
$\displaystyle+\sum\limits_{|j|=1}e_{(j)}f^{(j)}(\underline{z})$
$\displaystyle=$ $\displaystyle 2\int_{\Gamma}{\cal
D}_{\underline{z}}E_{0}(\underline{y}-\underline{z})n(\underline{y})R(\underline{y},\underline{z})d\underline{y}-2\int_{\Gamma}{\cal
D}_{\underline{z}}E_{1}(\underline{y}-\underline{z})n(\underline{y}){\cal
D}_{\underline{y}}R(\underline{y},\underline{z})d\underline{y}$
$\displaystyle+\sum\limits_{|j|=1}e_{(j)}f^{(j)}(\underline{z})$
$\displaystyle=$ $\displaystyle
2\int_{\Gamma}E_{0}(\underline{y}-\underline{z})n(\underline{y}){\cal
D}_{\underline{y}}R(\underline{y},\underline{z})d\underline{y}+\sum\limits_{|j|=1}e_{(j)}f^{(j)}(\underline{z}).$

Note that if $\exists i:j_{i}>l_{i}$, then
$\partial_{\underline{y}}^{(j)}(\underline{y}-\underline{z})^{l}=0$. Thus,
since
$f^{(0)}(\underline{y})=f^{(0)}(\underline{z})+\displaystyle\sum\limits_{|l|=1}f^{(l)}(\underline{z})(\underline{y}-\underline{z})^{l}+R(\underline{y},\underline{z})$,
we have

$\displaystyle{\cal
D}_{\underline{y}}R(\underline{y},\underline{z})=\sum\limits_{|j|=1}e_{(j)}\partial_{\underline{y}}^{(j)}\Big{[}\tilde{f}(\underline{y})-\tilde{f}(\underline{z})-\sum\limits_{|l|=1}\partial_{\underline{z}}^{(l)}\tilde{f}(\underline{z})(\underline{y}-\underline{z})^{l}\Big{]}=\sum\limits_{|j|=1}e_{(j)}\partial_{\underline{y}}^{(j)}\tilde{f}(\underline{y})-\sum\limits_{|j|=1}e_{(j)}\partial_{\underline{z}}^{(j)}\tilde{f}(\underline{z}).$

When this is substituted in (13) we get

$\displaystyle\sum\limits_{|j|=1}e_{(j)}[\mathcal{S}_{1}f]^{(j)}(\underline{z})$
$\displaystyle=$ $\displaystyle
2\int_{\Gamma}E_{0}(\underline{y}-\underline{z})n(\underline{y})\Bigg{\\{}\sum\limits_{|j|=1}e_{(j)}\partial_{\underline{y}}^{(j)}\tilde{f}(\underline{y})-\sum\limits_{|j|=1}e_{(j)}\partial_{\underline{z}}^{(j)}\tilde{f}(\underline{z})\Bigg{\\}}+\sum\limits_{|j|=1}e_{(j)}f^{(j)}(\underline{z}).$

Then (12) now follows from the above, since
$\displaystyle\int_{\Gamma}E_{0}(\underline{\zeta}-\underline{y})n(\underline{\zeta})d\underline{\zeta}=\frac{1}{2}$
when $\underline{y},\underline{\zeta}\in\Gamma$.

Let us prove (11) for $j=0.$

$\displaystyle[\mathcal{S}_{1}^{2}f]^{(0)}(\underline{z})$ $\displaystyle=$
$\displaystyle
2\sum\limits_{s=0}^{1}\int_{\Gamma}(-1)^{s}E_{s}(\underline{y}-\underline{z})n(\underline{y}){\cal
D}_{\underline{y}}^{s}\widetilde{\mathcal{S}_{k}f}(\underline{y})d\underline{y}$
$\displaystyle=$ $\displaystyle
2\int_{\Gamma}E_{0}(\underline{y}-\underline{z})n(\underline{y})\widetilde{\mathcal{S}_{1}f}(\underline{y})d\underline{y}-2\int_{\Gamma}E_{1}(\underline{y}-\underline{z})n(\underline{y}){\cal
D}_{\underline{y}}\widetilde{\mathcal{S}_{1}f}(\underline{y})d\underline{y}$
$\displaystyle=$ $\displaystyle
2\int_{\Gamma}E_{0}(\underline{y}-\underline{z})n(\underline{y})\widetilde{\mathcal{S}_{1}f}(\underline{y})d\underline{y}-2\int_{\Gamma}E_{1}(\underline{y}-\underline{z})n(\underline{y})\sum\limits_{|j|=1}e_{(j)}\partial^{(j)}_{\underline{y}}\widetilde{\mathcal{S}_{1}f}(\underline{y})d\underline{y}.$

On substituting (9) and (12) together with the observation that
$\partial_{\underline{y}}^{(j)}\widetilde{\mathcal{S}_{1}f}|_{\Gamma}=[\mathcal{S}_{1}f]^{(j)}$
we obtain

$\displaystyle[\mathcal{S}_{k}^{2}f]^{(0)}(\underline{z})$ $\displaystyle=$
$\displaystyle
2\int_{\Gamma}E_{0}(\underline{y}-\underline{z})n(\underline{y})[\mathcal{S}_{k}f]^{(0)}(\underline{y})d\underline{y}-2\int_{\Gamma}E_{1}(\underline{y}-\underline{z})n(\underline{y})\sum\limits_{|j|=1}e_{(j)}[\mathcal{S}_{1}f]^{(j)}(\underline{y})d\underline{y}$
$\displaystyle=$ $\displaystyle
2\int_{\Gamma}E_{0}(\underline{y}-\underline{z})n(\underline{y})\Big{[}2\int_{\Gamma}E_{0}(\underline{\zeta}-\underline{y})n(\underline{\zeta})\tilde{f}(\underline{\zeta})d\underline{\zeta}-2\int_{\Gamma}E_{1}(\underline{\zeta}-\underline{y})n(\underline{\zeta}){\cal
D}_{\underline{\zeta}}\tilde{f}(\underline{\zeta})d\underline{\zeta}\Big{]}d\underline{y}$
$\displaystyle-2\int_{\Gamma}E_{1}(\underline{y}-\underline{z})n(\underline{y})\Big{[}2\int_{\Gamma}E_{0}(\underline{\zeta}-\underline{y})n(\underline{\zeta})\Big{(}\sum\limits_{|j|=1}e_{(j)}f^{(j)}(\underline{\zeta})\Big{)}d\underline{\zeta}\Big{]}d\underline{y}.$

Since
$2\int_{\Gamma}E_{0}(\underline{y}-\underline{z})n(\underline{y})\Big{[}2\int_{\Gamma}E_{0}(\underline{\zeta}-\underline{y})n(\underline{\zeta})\tilde{f}(\underline{\zeta})d\underline{\zeta}\Big{]}d\underline{y}=f^{(0)}(\underline{z})$
according to the classical involution property and
$\sum_{|j|=1}e_{(j)}f^{(j)}(\underline{y})=\sum_{|j|=1}e_{(j)}\partial_{\underline{y}}^{(j)}\tilde{f}(\underline{y})={\cal
D}_{\underline{y}}\tilde{f}(\underline{y}),$ it is therefore enough to show
that the following expression vanishes

$\displaystyle Q$ $\displaystyle:=$ $\displaystyle
2\int_{\Gamma}E_{0}(\underline{y}-\underline{z})n(\underline{y})\Big{[}2\int_{\Gamma}-E_{1}(\underline{\zeta}-\underline{y})n(\underline{\zeta}){\cal
D}_{\underline{\zeta}}\tilde{f}(\underline{\zeta})d\underline{\zeta}\Big{]}d\underline{y}-2\int_{\Gamma}E_{1}(\underline{y}-\underline{z})n(\underline{y})\Big{[}2\int_{\Gamma}E_{0}(\underline{\zeta}-\underline{y})n(\underline{\zeta}){\cal
D}_{\underline{\zeta}}\tilde{f}(\underline{\zeta})d\underline{\zeta}\Big{]}d\underline{y}$
$\displaystyle=$ $\displaystyle
2\sum\limits_{s=0}^{1}\int_{\Gamma}(-1)^{s}E_{s}(\underline{y}-\underline{z})n(\underline{y})\Big{[}2\int_{\Gamma}-{\cal
D}_{\underline{y}}^{s}E_{1}(\underline{\zeta}-\underline{y})n(\underline{\zeta}){\cal
D}_{\underline{\zeta}}\tilde{f}(\underline{\zeta})d\underline{\zeta}\Big{]}d\underline{y}.$

Fubini’s theorem leads to

$\displaystyle Q$ $\displaystyle=$ $\displaystyle
4\int_{\Gamma}\int_{\Gamma}\sum\limits_{s=0}^{1}(-1)^{s}E_{s}(\underline{y}-\underline{z})n(\underline{y})\Big{[}-{\cal
D}_{\underline{y}}^{s}E_{1}(\underline{\zeta}-\underline{y})d\underline{y}\Big{]}n(\underline{\zeta}){\cal
D}_{\underline{\zeta}}d\underline{\zeta}.$

Define $F(\underline{y}):=-E_{1}(\underline{\zeta}-\underline{y})$. It
suffices to show that

$\displaystyle\sum\limits_{s=0}^{1}\int_{\Gamma}(-1)^{s}E_{s}(\underline{y}-\underline{z})n(\underline{y}){\cal
D}_{\underline{y}}^{s}F(\underline{y})d\underline{y}=0.$

Let
$\Gamma^{*}=\Gamma\setminus(\Gamma_{\epsilon}(\underline{z})\cup\Gamma_{\epsilon}(\underline{\zeta}))\cup-C^{*}_{\epsilon}(\underline{z})\cup-C^{*}_{\epsilon}(\underline{\zeta})$
and $\Omega_{*}$ its interior. Clearly, $F(\underline{y})$ is bimonogenic on
$\Omega_{*}$ and continuous on $\Omega_{*}\cup\Gamma^{*}$, so we can apply
Cauchy’s formula (for
$\underline{z}\in\mathbb{R}^{m}\setminus\overline{\Omega}_{*}$) to get

$\displaystyle\sum\limits_{s=0}^{1}\int_{\Gamma^{*}}(-1)^{s}E_{s}(\underline{y}-\underline{z})n(\underline{y}){\cal
D}_{\underline{y}}^{s}F(\underline{y})d\underline{y}=0.$

Then we are reduced to proving that

$\displaystyle\lim\limits_{\epsilon\to
0}\Bigg{\\{}\sum\limits_{s=0}^{1}\int_{C^{*}_{\epsilon}(\underline{z})\cup
C^{*}_{\epsilon}(\underline{\zeta})}(-1)^{s}E_{s}(\underline{y}-\underline{z})n(\underline{y}){\cal
D}_{\underline{y}}^{s}F(\underline{y})d\underline{y}\Bigg{\\}}=0.$ (14)

We compute

$\displaystyle\int_{C^{*}_{\epsilon}(\underline{z})\cup
C^{*}_{\epsilon}(\underline{\zeta})}\bigg{[}E_{0}(\underline{y}-\underline{z})n(\underline{y})E_{1}(\underline{\zeta}-\underline{y})+E_{1}(\underline{y}-\underline{z})n(\underline{y})E_{0}(\underline{\zeta}-\underline{y})\bigg{]}d\underline{y}=$
(15) $\displaystyle=$ $\displaystyle
c_{m}\Biggl{\\{}\int_{C^{*}_{\epsilon}(\underline{z})}\frac{-(\underline{y}-\underline{z})}{|\underline{y}-\underline{z}|^{m}}\frac{\underline{y}-\underline{z}}{|\underline{y}-\underline{z}|}\frac{1}{|\underline{\zeta}-\underline{y}|^{m-2}}d\underline{y}+\int_{C^{*}_{\epsilon}(\underline{z})}\frac{1}{|\underline{y}-\underline{z}|^{m-2}}\frac{\underline{y}-\underline{z}}{|\underline{y}-\underline{z}|}\frac{-(\underline{\zeta}-\underline{y})}{|\underline{\zeta}-\underline{y}|^{m}}d\underline{y}$
$\displaystyle+\int_{C^{*}_{\epsilon}(\underline{\zeta})}\frac{-(\underline{y}-\underline{z})}{|\underline{y}-\underline{z}|^{m}}\frac{\underline{y}-\underline{\zeta}}{|\underline{y}-\underline{\zeta}|}\frac{1}{|\underline{\zeta}-\underline{y}|^{m-2}}d\underline{y}+\int_{C^{*}_{\epsilon}(\underline{\zeta})}\frac{1}{|\underline{y}-\underline{z}|^{m-2}}\frac{\underline{y}-\underline{\zeta}}{|\underline{y}-\underline{\zeta}|}\frac{-(\underline{\zeta}-\underline{y})}{|\underline{\zeta}-\underline{y}|^{m}}d\underline{y}\Biggr{\\}}$
$\displaystyle=$
$\displaystyle\frac{c_{m}}{\epsilon^{m-1}}\Bigg{\\{}\int_{C^{*}_{\epsilon}(\underline{z})}\frac{d\underline{y}}{|\underline{\zeta}-\underline{y}|^{m-2}}-\int_{C^{*}_{\epsilon}(\underline{\zeta})}\frac{d\underline{y}}{|\underline{y}-\underline{z}|^{m-2}}$
$\displaystyle+\int_{C^{*}_{\epsilon}(\underline{z})}\frac{(\underline{y}-\underline{z})(\underline{y}-\underline{\zeta})}{|\underline{\zeta}-\underline{y}|^{m}}d\underline{y}-\int_{C^{*}_{\epsilon}(\underline{\zeta})}\frac{(\underline{y}-\underline{z})(\underline{y}-\underline{\zeta})}{|\underline{y}-\underline{z}|^{m}}d\underline{y}\Bigg{\\}}.$

After the change of variable
$\underline{y}=\underline{z}-\underline{t}+\underline{\zeta}$, we get that the
first and second expressions in curly brackets are equal. This observation
applies to the third and fourth terms as well and
$[\mathcal{S}_{1}^{2}f]^{(0)}=f^{(0)}$ as claimed. Now that we have the above
claim, we set
$G=\\{[\mathcal{S}_{1}^{2}f]^{(j)}(\underline{z})-f^{(j)}(\underline{z}),\,|j|\leq
1\\}$. Clearly, $G^{(0)}=0$. By Plemelj-Privalov’s theorem [2],
$G\in{\mbox{Lip}}(1+\alpha,\Gamma)$ and according to Lemma 1 it suffices to
show $\sum\limits_{|j|=1}e_{(j)}G^{(j)}=0.$

On applying (12) twice, we obtain

$\displaystyle\sum\limits_{|j|=1}e_{(j)}G^{(j)}(\underline{z})$
$\displaystyle=$
$\displaystyle\sum\limits_{|j|=1}e_{(j)}\big{[}\mathcal{S}_{1}(\mathcal{S}_{1}f)\big{]}^{(j)}(\underline{z})-\sum\limits_{|j|=1}e_{(j)}f^{(j)}(\underline{z})$
$\displaystyle=$ $\displaystyle
2\int_{\Gamma}E_{0}(\underline{y}-\underline{z})n(\underline{y})\Bigg{(}\sum\limits_{|j|=1}e_{(j)}(\mathcal{S}_{1}f)^{(j)}(\underline{y})\Bigg{)}d\underline{y}-\sum\limits_{|j|=1}e_{(j)}f^{(j)}(\underline{z})$
$\displaystyle=$ $\displaystyle
2\int_{\Gamma}E_{0}(\underline{y}-\underline{z})n(\underline{y})\Bigg{\\{}2\int_{\Gamma}E_{0}(\underline{\zeta}-\underline{y})n(\underline{\zeta})\Big{(}\sum\limits_{|j|=1}e_{(j)}f^{(j)}(\underline{\zeta})\Big{)}d\underline{\zeta}\Bigg{\\}}d\underline{y}-\sum\limits_{|j|=1}e_{(j)}f^{(j)}(\underline{z}).$

By the involution property of the classical singular operator, we get
$\sum\limits_{|j|=1}e_{(j)}G^{(j)}(\underline{z})=0$, which completes the
proof. ∎

It is immediate that the operators
$\mathcal{P}^{+}=\frac{1}{2}(I+\mathcal{S}_{1})$ and
$\mathcal{P}^{-}=\frac{1}{2}(I-\mathcal{S}_{1})$ are projections on
${\mbox{Lip}}(1+\alpha,\Gamma)$, that is

$\mathcal{P}^{+}\mathcal{P}^{+}=\mathcal{P}^{+},\,\mathcal{P}^{-}\mathcal{P}^{-}=\mathcal{P}^{-},\,\mathcal{P}^{+}\mathcal{P}^{-}=0,\,\mathcal{P}^{-}\mathcal{P}^{+}=0.$

Consequently,
${\mbox{Lip}}(1+\alpha,\Gamma)={\mbox{Lip}}^{+}(1+\alpha,\Gamma)\oplus{\mbox{Lip}}^{-}(1+\alpha,\Gamma),$
where ${\mbox{Lip}}^{\pm}(1+\alpha,\Gamma):=\mbox{im}\mathcal{P}^{\pm}$.

######  Theorem 3.

The Whitney data $f\in\emph{{\mbox{Lip}}}(1+\alpha,\Gamma)$ belongs to
$\emph{{\mbox{Lip}}}^{+}(1+\alpha,\Gamma)$ if and only if there exists a
bimonogenic function $F$ in $\Omega_{+}$ which together with ${\cal
D}_{\underline{x}}F$ continuously extends to $\Gamma$ and such that

$F|_{\Gamma}=f^{(0)},\,\,{\cal
D}_{\underline{x}}F|_{\Gamma}=\sum\limits_{|j|=1}e_{(j)}f^{(j)}.$ (16)

###### Proof.

The proof proceeds along the same lines as the proof of [1, Theorem 6], but we
use (12) instead. Let us prove necessity. By definition, if
$f\in{\mbox{Lip}}^{+}(1+\alpha,\Gamma)$ there exists
$g\in{\mbox{Lip}}(1+\alpha,\Gamma)$ such that
$f=\frac{1}{2}(g+\mathcal{S}_{1}g)$, i.e.
$f^{(j)}(\underline{x})=\frac{1}{2}[I^{(j)}+\mathcal{S}_{1}^{(j)}]g,|j|\leq
1$, where $I^{(j)}g=g^{(j)}$ is the identity operator. Let us introduce the
function $F$ given by the Cauchy type transform
$F(\underline{x})=[\mathcal{C}_{1}g]^{(0)}(\underline{x}),\,\underline{x}\in\Omega_{+}$,
which is clearly bimonogenic in $\Omega_{+}$. Thus, for
$\underline{z}\in\Gamma$ we get from (10) that

$F(\underline{z})=\lim_{\underline{x}\to\underline{z}\atop\underline{x}\in\Omega_{+}}F(\underline{x})=\big{(}[\mathcal{C}_{1}g]^{(0)}\big{)}^{+}(\underline{z})=\frac{1}{2}[I^{(0)}+\mathcal{S}_{1}^{(0)}]g=f^{(0)}(\underline{z}).$

On the other hand,

${\cal D}_{\underline{x}}F(\underline{x})={\cal
D}_{\underline{x}}\Big{[}\int_{\Gamma}E_{0}(\underline{y}-\underline{x})n(\underline{y})\tilde{g}(\underline{y})d\underline{y}-\int_{\Gamma}E_{1}(\underline{y}-\underline{x})n(\underline{y}){\cal
D}_{\underline{y}}\tilde{g}(\underline{y})d\underline{y}\Big{]}=\mathcal{C}_{0}[{\cal
D}\tilde{g}](\underline{x})=\mathcal{C}_{0}\Big{[}\sum\limits_{|j|=1}e_{(j)}\partial_{\underline{x}}^{(j)}\tilde{g}\Big{]}(\underline{x}).$

From (3) we have for $\underline{z}\in\Gamma$

$\displaystyle[{\cal
D}_{\underline{x}}F](\underline{z})=\lim_{\underline{x}\to\underline{z}\atop\underline{x}\in\Omega_{+}}{\cal
D}_{\underline{x}}F(\underline{x})=\Big{[}\mathcal{C}_{0}({\cal
D}_{\underline{x}}\tilde{g})\Big{]}^{+}(\underline{z})$ $\displaystyle=$
$\displaystyle\frac{1}{2}[I+\mathcal{S}_{0}]\Big{(}\sum\limits_{|j|=1}e_{(j)}g^{(j)}\Big{)}(\underline{z})$
$\displaystyle=$
$\displaystyle\frac{1}{2}\Big{[}\sum\limits_{|j|=1}e_{(j)}g^{(j)}(\underline{z})+2\int_{\Gamma}E_{0}(\underline{y}-\underline{z})n(\underline{y})\Big{(}\sum\limits_{|j|=1}e_{(j)}g^{(j)}(\underline{y})\Big{)}d\underline{y}\Big{]}.$

Hence it follows from (12) that ${\cal
D}_{\underline{x}}F|_{\Gamma}=\sum\limits_{|j|=1}e_{(j)}f^{(j)}(\underline{z}).$

We now prove sufficiency. Assume there exists such bimonogenic function $F$
satisfying (16). It turns out that nothing more need be done to prove that
$[\mathcal{P}^{+}f]^{(j)}=f^{(j)}$ for all $0\leq|j|\leq 1$. Indeed, in that
case, $f\in\mbox{im}\mathcal{P}^{+}$ as claimed. Combining our assumptions
with (7) gives

$\displaystyle F(\underline{x})$ $\displaystyle=$
$\displaystyle\int_{\Gamma}E_{0}(\underline{y}-\underline{x})n(\underline{y})F(\underline{y})d\underline{y}-\int_{\Gamma}E_{1}(\underline{y}-\underline{x})n(\underline{y}){\cal
D}_{\underline{y}}F(\underline{y})d\underline{y},\,\,\underline{x}\in\Omega_{+}$
$\displaystyle=$
$\displaystyle\int_{\Gamma}E_{0}(\underline{y}-\underline{x})n(\underline{y})f^{(0)}(\underline{y})d\underline{y}-\int_{\Gamma}E_{1}(\underline{y}-\underline{x})n(\underline{y})\Big{(}\sum\limits_{|j|=1}e_{(j)}f^{(j)}(\underline{y})\Big{)}d\underline{y}=[\mathcal{C}_{1}f]^{(0)}(\underline{x}).$

It is easy to see from this and (10) that

$\displaystyle[\mathcal{P}^{+}f]^{(0)}(\underline{z}):=\frac{1}{2}[I^{(0)}+\mathcal{S}_{1}^{(0)}]f(\underline{z})=\big{(}[\mathcal{C}_{1}f]^{(0)}\big{)}^{+}(\underline{z})=F(\underline{z})=f^{(0)}(\underline{z}).$

On the other hand,

$\displaystyle\sum\limits_{|j|=1}e_{(j)}[\mathcal{P}^{+}f]^{(j)}(\underline{z})$
$\displaystyle:=$
$\displaystyle\sum\limits_{|j|=1}e_{(j)}\frac{1}{2}[I^{(j)}+\mathcal{S}_{1}^{(j)}]f(\underline{z})$
(17) $\displaystyle=$
$\displaystyle\frac{1}{2}\Big{[}\sum\limits_{|j|=1}e_{(j)}f^{(j)}(\underline{z})+\sum\limits_{|j|=1}e_{(j)}[\mathcal{S}_{1}f]^{(j)}(\underline{z})\Big{]}$
$\displaystyle=$
$\displaystyle\frac{1}{2}\Big{[}\sum\limits_{|j|=1}e_{(j)}f^{(j)}(\underline{z})+2\int_{\Gamma}E_{0}(\underline{y}-\underline{z})n(\underline{y})\Big{(}\sum\limits_{|j|=1}e_{(j)}f^{(j)}(\underline{y})\Big{)}d\underline{y}\Big{]}$

the last equality being a consequence of (12). Since the density function in
(17) satisfies the Hölder condition on $\Gamma$ and due to (16) it represents
the interior limiting value of the monogenic function ${\cal
D}_{\underline{x}}F$ then (3) yields

$\sum\limits_{|j|=1}e_{(j)}f^{(j)}(\underline{z})=\Big{[}\mathcal{C}_{0}\big{(}\sum\limits_{|j|=1}e_{(j)}f^{(j)}\big{)}\Big{]}^{+}(\underline{z})=\frac{1}{2}\big{[}I+\mathcal{S}_{0}\big{]}\big{(}\sum\limits_{|j|=1}e_{(j)}f^{(j)}\big{)}(\underline{z}),$

or equivalently

$\frac{1}{2}\Big{[}\sum\limits_{|j|=1}e_{(j)}f^{(j)}(\underline{z})\Big{]}=\mathcal{S}_{0}\Big{(}\sum\limits_{|j|=1}e_{(j)}f^{(j)}\Big{)}(\underline{z}),$

which gives

$\sum\limits_{|j|=1}e_{(j)}[\mathcal{P}^{+}f]^{(j)}(\underline{z})=\sum\limits_{|j|=1}e_{(j)}f^{(j)}(\underline{z})$

when substituted in (17) and hence the assertion follows from Lemma 1. ∎

Similar arguments to those above, but using (8) instead, show the following
theorem and its proof is omitted.

######  Theorem 4.

The Whitney data $f\in\emph{{\mbox{Lip}}}(1+\alpha,\Gamma)$ belongs to
$\emph{{\mbox{Lip}}}^{-}(1+\alpha,\Gamma)$ if and only if there exists a
bimonogenic function $F$ in $\Omega_{-}$ vanishing at infinity satisfying
${\cal
D}_{\underline{x}}F=\scriptscriptstyle\mathcal{O}\big{(}\displaystyle\frac{1}{|\underline{x}|}\big{)}$
as $\underline{x}\to\infty$, which together with ${\cal D}_{\underline{x}}F$
continuously extends to $\Gamma$ and such that

$F|_{\Gamma}=f^{(0)},\,\,{\cal
D}_{\underline{x}}F|_{\Gamma}=\sum\limits_{|j|=1}e_{(j)}f^{(j)}.$

In other words, given $f\in{\mbox{Lip}}(1+\alpha,\Gamma)$, the problem of
finding a sectionally bimonogenic $F$ satisfying the boundary conditions

$\begin{cases}F^{+}(\underline{z})-F^{-}(\underline{z})=f^{(0)}(\underline{z}),&\,\underline{z}\in\Gamma\\\
[{\cal D}_{\underline{x}}F]^{+}(\underline{z})-[{\cal
D}_{\underline{x}}F]^{-}(\underline{z})=\sum\limits_{|j|=1}e_{(j)}f^{(j)}(\underline{z}),&\,\underline{z}\in\Gamma\\\
F(\infty)=0,\,\,{\cal
D}_{\underline{x}}F=\scriptscriptstyle\mathcal{O}\big{(}\dfrac{1}{|\underline{x}|}\big{)},&{\mbox{as}}\,\,\underline{x}\to\infty\end{cases}$

has a unique solution and it is expressed by $F=[\mathcal{C}_{1}f]^{(0)}.$

##  5 Concluding remarks

In the case of considering $k$-monogenic functions with $k>1$, a general
representation formula allowed us in [2] to define the associated Cauchy
transform and the singular integral operator $\mathcal{S}_{k}$ related to
Lipschitz functions of arbitrary order ${\mbox{Lip}}(k+\alpha,\Gamma)$. We can
ask whether Theorems 2 \- 4 are still valid for polymonogenic functions and
Lipschitz classes of arbitrary order. Partial evidence support the belief that
the method developed in this paper can also be successfully applied to the
general case. The confirmation of this hypothesis inspires further analysis
and research, which will be the subject of future work.

##  6 Acknowledgments

L. De la Cruz Toranzo was supported by a Research Fellowship under the
auspices of the Alexander von Humboldt Foundation.

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