1 Introduction
In some way, there are two branches of Clifford analysis. The first one is the real Clifford analysis introduced by Brack, Delanghe, and Sommen in [1] which studied function theory with values in a real Clifford algebra defined on a nonempty subset of the Euclidean space \(R^{n+1} \). Many important theoretic results, such as the Cauchy integral formula, the Cauchy theorem, the Taylor and the Laurent series expansion, the Liouville theorem, and the Morera theorem, have been obtained, and they are the extensions of the well-known classical theorems in one complex variable. Beyond these, a lot of scholars have studied many properties of function theory in the real Clifford analysis. Eriksson and Leutwiler [2‐5] introduced the hypermonogenic function and studied some properties of it. Huang [6], Qiao [7‐9], Xie [10‐12], and Yang [13‐15] obtained many results in Clifford analysis.
The second one is the complex Clifford analysis. In the early 1990s, Ryan [16‐19] introduced the definition of the complex regular function and obtained the Cauchy integral formula whose method is similar to the classical function with one complex variable. In recent years, Ku, Du [20, 21] obtained some properties of complex regular functions using the isotonic function.
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Based on the above theoretical study and practical background, we construct an analogue of Bochner–Martinelli kernel in several complex variables. We first define a Teodorescu operator with B-M kernel in the complex Clifford analysis and prove the boundedness of this operator. Then we give an inequality similar to the classical Hile lemma about real vector which plays a key role in the following proof. Finally, we prove the Hölder continuity and γ-integrability of this operator.
2 Preliminaries
Let \(\mathrm{Cl}_{0,n}(C)\) be a complex Clifford algebra over n+1-dimensional Euclidean space \({\mathbf{C}}^{n+1}\). \(\mathrm{Cl}_{0,n}(C)\) has the basis \(e_{0}, e_{1}, e_{2}, \ldots, e_{n}; e_{1}e_{2}, e_{1}e_{3}, \ldots, e_{1}e_{n}; e_{2}e_{3}, \ldots, e_{2}e_{n}; \ldots; e_{n-1}e_{n}; \ldots; [4] e_{1}\cdots{e_{n}}\). Hence, an arbitrary element of the basis may be written as \(e_{A}=e_{\alpha_{1}}\cdots e_{\alpha_{h}}\), where \(A=\{\alpha_{1}, \ldots, \alpha_{h} \}\subseteq\{1, \ldots, n\}\), \(1\le\alpha_{1}<\alpha_{2}<\cdots<\alpha_{h}\le n\) and when \(A=\emptyset\), \(e_{A}=e_{0}=1\). So, the complex Clifford algebra is composed of elements having the type \(a=\sum_{A}z_{A}e_{A}\), where \(z_{A}\) are complex numbers.
The basis in Clifford algebra satisfies
$${e_{i}^{2}=-1,\quad i=1, 2, \ldots, n}, \qquad e_{i}e_{j}=-e_{j}e_{i},\quad 1\leq i< j\leq n, (i\neq j). $$
Define the norm of Clifford numbers as follows:
$$\biggl\vert \sum_{A}z_{A}e_{A} \biggr\vert =\sqrt{(a, a)}= \biggl(\sum_{A} \vert z_{A} \vert ^{2} \biggr)^{\frac{1}{2}}. $$
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Let \(\Omega\subset\mathbf{C}^{n+1}\) be an open connected nonempty set. Then the function which is defined on Ω and valued in \(\mathrm{Cl}_{0,n}(C)\) can be expressed as \(f(z)=\sum_{A}{f_{A}(z)e_{A}}\), where \(f_{A}(z)\) are complex-valued functions. Let
$${F^{(r)}_{\Omega}=\biggl\{ f\Big|f:\Omega\rightarrow \mathrm{Cl}_{0,n}(C), f(z)=\sum_{A}{f_{A}(z)e_{A}}, f_{A}(z)\in C^{r}(\Omega)\biggr\} }. $$
Dirac operators are introduced as follows [6]:
$$\begin{gathered} D^{l} f=\sum_{i=0}^{n}e_{i} \frac{\partial f}{\partial z_{i}};\qquad \overline{D^{l}}f=e_{0}{ \frac{\partial f }{\partial z_{0}}}-\sum_{i=1}^{n}e_{i}{ \frac{\partial f}{\partial z_{i}}}; \\ D^{r} f=\sum_{i=0}^{n} \frac{\partial f}{\partial z_{i}}e_{i};\qquad \overline{D^{r}}f={ \frac{\partial f }{\partial z_{0}}}e_{0}-\sum_{i=1}^{n}{ \frac{\partial f}{\partial z_{i}}}e_{i}. \end{gathered} $$
Definition 2.1
([16])
If \(\Omega\subset C^{n+1}\), \(f: \Omega\rightarrow \mathrm{Cl}_{0,n}(C)\) satisfies: then \(f(z)\) is called a complex left regular function on Ω.
(1)
\(f_{A}(z)\) is a holomorphic function for any \(z_{j}\in\Omega\),
(2)
\(D_{l}f(z)=0\), \(\forall z\in\Omega\),
Definition 2.2
([16])
If \(\Omega\subset C^{n+1}\), \(f: \Omega\rightarrow \mathrm{Cl}_{0,n}(C)\) satisfies: then \(f(z)\) is called a complex right regular function on Ω.
(1)
\(f_{A}(z)\) is a holomorphic function for any \(z_{j}\in\Omega\),
(2)
\(D_{r}f(z)=0\), \(\forall z\in\Omega\),
Lemma 2.1
(Hadamard lemma [22])
Let
\(\Omega\subset R^{n+1}\)
be a bounded domain, \(n\geq2\). If
α, β
satisfy
\(0<\alpha,\beta <n+1\), and
\(\alpha+\beta>n+1\), then for any
\(x_{1},x_{2}\in R^{n+1}\), \(x_{1}\neq x_{2}\), we have
where
\(J_{1}\)
is a positive constant related to
α
and
β.
$$\int_{\Omega} \vert t-x_{1} \vert ^{-\alpha} \vert t-x_{2} \vert ^{-\beta}\,dt\leq J_{1} \vert x_{1}-x_{2} \vert ^{(n+1)-\alpha-\beta}, $$
Lemma 2.2
([22])
Let
\(\Omega\subset R^{n+1}\)
be a bounded domain, when
\(\alpha< n+1\), for any
\(y\in R^{n+1}\), we have
where
M
is a positive constant only related to
α
and the size of Ω.
$$\int_{\Omega} \vert x-y \vert ^{-\alpha}\,dx\leq M, $$
Lemma 2.3
(Hölder inequality [23])
If
\(f_{k}\in L^{p_{k}}(\Omega)\), \(k=1,2,\ldots,n\), and
then
\({f_{1}f_{2}\cdots f_{n}\in L^{p}(\Omega)}\), and
$${\frac{1}{p}=\frac{1}{p_{1}}+\frac{1}{p_{2}}+\cdots+\frac{1}{p_{n}} \leq1,} $$
$$L^{p}(f_{1}f_{2}\cdots f_{n})\leq L^{p_{1}}(f_{1})L^{p_{2}}(f_{2}) \cdots L^{p_{n}}(f_{n}), \quad p\geq1. $$
Lemma 2.4
(Minkowski inequality [23])
If
\({f_{1},f_{2},\ldots,f_{n}\in L^{p}(\Omega)}\), then
\(f_{1}+f_{2}+\cdots+f_{n}\in L^{p}(\Omega)\), and
$${L^{p}(f_{1}+f_{2}+\cdots+f_{n})\leq L^{p}(f_{1})+L^{p}(f_{2})+ \cdots+L^{p}(f_{n}), p\geq1.} $$
Lemma 2.5
([23])
Let
\(L^{p}(\Omega,\mathrm{Cl}_{0,n}(R))\)
represent the set of all
p
order integrable functions which are defined on the bounded domain
\(\Omega \subset R^{n+1}\), and with values in the real Clifford algebra
\(\mathrm{Cl}_{0,n}(R)\), define the norm of
φ
as follows:
when
\(1\leq r \leq p\),
is true.
$$\Vert \varphi \Vert _{\Omega,p}= \biggl( \int_{\Omega} \bigl\vert \varphi (x) \bigr\vert ^{p}\,dV_{x} \biggr)^{\frac{1}{p}},\quad p\geq1, $$
$$L^{p}\bigl(\Omega, \mathrm{Cl}_{0,n}(R)\bigr)\subset L^{r} \bigl(\Omega, \mathrm{Cl}_{0,n}(R)\bigr) $$
The notations used in this paper are as follows:
(1)
\(\omega_{2n+2}\) represents the surface area of unit sphere in a \(2n+2\)-dimensional real Euclidean space.
(2)
\(M_{i}\)
\(\{i=1,2,3\}\), \(K_{i}\)
\(\{i=1,\dots,16\}\) are constants only related to n and the size of domain Ω in this paper.
(3)
\(dV_{\xi}=d\zeta_{0}\wedge d\zeta_{1}\wedge\cdots\wedge d\zeta_{n}\wedge d\eta_{0}\wedge d\eta_{1}\wedge\cdots\wedge d\eta _{n}\), \(\zeta_{j}\in R\), \(\eta_{j}\in R\), (\(j=0, 1, \ldots, n\)), \(\xi_{j}=\zeta _{j}+i\eta_{j}\).
(4)
\(d\bar{\xi}\wedge d{\xi}=d\bar{\xi}_{0}\wedge d\bar{\xi}_{1}\wedge \cdots d\bar{\xi}_{n}\wedge d{\xi_{0}}\wedge d{\xi_{1}}\dots\wedge d{\xi_{n}}\).
(5)
\(d\bar{\xi}\wedge d{\xi}=(2i)^{n+1}\,dV_{\xi}\).
3 Some properties of a T operator with B-M kernel in the complex Clifford analysis
In this section, we discuss some properties of a singular integral operator.
Definition 3.1
Let \(\Omega\subset C^{n+1}\) be a bounded domain, \(\varphi\in L^{p}(\overline{\Omega},\mathrm{Cl}_{0,n}(C))\), \(z\in C^{n+1}\), then
is called T operator with B-M kernel.
$$\begin{aligned} &(T\varphi) (z) \\ &\quad=\frac{1}{\omega_{2n+2}(2i)^{n+1}} \int_{\Omega}\varphi(\xi) \biggl(\frac{\sum_{k=0}^{n}(\xi_{k}-z_{k})\bar{e}_{k}}{ \vert \xi-z \vert ^{2n+2}}+ \frac{\sum_{k=0}^{n}(\overline{\xi_{k}-z_{k}})\bar {e}_{k}}{ \vert \xi-z \vert ^{2n+2}} \biggr)\,d\bar{\xi}\wedge d{\xi} \end{aligned}$$
Theorem 3.1
Let
\(\Omega\subset C^{n+1}\)
be a bounded domain, \(\varphi\in L^{p}(\overline{\Omega},\mathrm{Cl}_{0,n}(C))\), \(p>n+1\), then
T
is bounded on
\(L^{p}(\Omega)\), and
$$\begin{aligned} \Vert T\varphi \Vert _{\Omega,p}\leq M_{1} \Vert \varphi \Vert _{\Omega,p}. \end{aligned}$$
(1)
Proof
Choose \(q>1\) such that \(\frac{1}{p}+\frac{1}{q}=1\), when \(p>2(n+1)\), we have \(1< q<\frac{2(n+1)}{2n+1}\), using Hölder’s inequality, we have
Because \(1< q<\frac{2(n+1)}{2n+1}\), we have \((2n+1)q<2(n+1)\). Using Lemma 2.2, for \(\forall z\in\Omega\), we have
So we have
Hence,
Let \(M_{1}=K_{4}K_{5} (\int_{\Omega}\,dV_{z} )^{\frac{1}{p}}\), we have
□
$$\begin{aligned} & \bigl\vert T\varphi(z) \bigr\vert \\ &\quad= \frac{1}{2^{n+1}\omega_{2n+2}} \biggl\vert \int_{\Omega}\varphi(\xi ) \biggl(\frac{\sum_{k=0}^{n}(\xi_{k}-z_{k})\bar{e}_{k}}{ \vert \xi-z \vert ^{2n+2}}+ \frac{\sum_{k=0}^{n}(\overline{\xi_{k}-z_{k}})\bar {e}_{k}}{ \vert \xi-z \vert ^{2n+2}} \biggr)\,d\bar{\xi}\wedge d{\xi} \biggr\vert \\ &\quad\leq\frac{K_{1}}{\omega_{2n+2}} \biggl( \int_{\Omega} \biggl\vert \varphi (\xi)\frac{\sum_{k=0}^{n}(\xi_{k}-z_{k})\bar{e}_{k}}{ \vert \xi -z \vert ^{2n+2}} \biggr\vert \,dV_{\xi} + \int_{\Omega} \biggl\vert \varphi(\xi)\frac{\sum_{k=0}^{n}(\overline{\xi _{k}-z_{k}})\bar{e}_{k}}{ \vert \xi-z \vert ^{2n+2}} \biggr\vert \,dV_{\xi} \biggr) \\ &\quad\leq K_{2} \biggl( \int_{\Omega} \bigl\vert \varphi(\xi) \bigr\vert \frac{1}{ \vert \xi -z \vert ^{2n+1}}\,dV_{\xi}+ \int_{\Omega} \bigl\vert \varphi(\xi) \bigr\vert \frac{1}{ \vert \xi-z \vert ^{2n+1}} \,dV_{\xi} \biggr) \\ &\quad \leq K_{3} \int_{\Omega} \bigl\vert \varphi(\xi) \bigr\vert \frac{1}{ \vert \xi-z \vert ^{2n+1}}\,dV_{\xi} \\ &\quad\leq K_{4} \Vert \varphi \Vert _{\Omega,p} \biggl( \int_{\Omega} \vert \xi -z \vert ^{-(2n+1)q}\,dV_{\xi} \biggr)^{\frac{1}{q}}. \end{aligned}$$
$$\int_{\Omega} \vert \xi-z \vert ^{-(2n+1)q}\,dV_{\xi} \leq K_{5}. $$
$$\bigl\vert T\varphi(z) \bigr\vert \leq K_{4}K_{5} \Vert \varphi \Vert _{\Omega,p}. $$
$$\biggl( \int_{\Omega} \bigl\vert T\varphi(z) \bigr\vert ^{p}\,dV_{z} \biggr)^{\frac{1}{p}}\leq K_{4}K_{5} \biggl( \int_{\Omega} \Vert \varphi \Vert _{\Omega,p}^{p}\,dV_{z} \biggr)^{\frac{1}{p}}. $$
$$\bigl\Vert T\varphi(z) \bigr\Vert _{\Omega,p}\leq M_{1} \Vert \varphi \Vert _{\Omega,p}. $$
Theorem 3.2
Let
\(z=z_{0}e_{0}+z_{1}e_{1}+z_{2}e_{2}+\cdots+z_{n}e_{n}\), \(\xi=\xi _{0}e_{0}+\xi_{1}e_{1}+\xi_{2}e_{2}+\cdots+\xi_{n}e_{n}\in\mathrm{Cl}_{0,n}(C)\), \(z\neq0\), \(\xi\neq0\), and
\(|z|\neq|\xi|\), n (≥2), m (≥0) be integers, then for any
i, \(0\leq i\leq n\), we have
where
$$\begin{aligned} \biggl\vert \frac{z_{i}}{ \vert z \vert ^{m+2}}-\frac{\xi_{i}}{ \vert \xi \vert ^{m+2}} \biggr\vert \leq \frac{ \vert z-\xi \vert [P_{m}(z,\xi)+ \vert z \vert ^{\frac{m}{2}} \vert \xi \vert ^{\frac{m}{2}} ]}{ \vert z \vert ^{m+1} \vert \xi \vert ^{m+1}}, \end{aligned}$$
(2)
$$P_{m}(z,\xi)= \sum_{k=0}^{m} \vert z \vert ^{m-k} \vert \xi \vert ^{k}={ \frac{ \vert z \vert ^{m+1}- \vert \xi \vert ^{m+1}}{ \vert z \vert - \vert \xi \vert }}. $$
Proof
Suppose \(|z|\leq|\xi|\) and insert the term \(z_{i}|z|^{m+2}\) in the following formula, then we have
$$\begin{aligned} & \biggl\vert \frac{z_{i}}{ \vert z \vert ^{m+2}}-\frac{\xi_{i}}{ \vert \xi \vert ^{m+2}} \biggr\vert \\ &\quad= \biggl\vert \frac{z_{i} \vert \xi \vert ^{m+2}-\xi_{i} \vert z \vert ^{m+2}}{ \vert z \vert ^{m+2} \vert \xi \vert ^{m+2}} \biggr\vert \\ &\quad= \biggl\vert \frac{z_{i} \vert \xi \vert ^{m+2}-z_{i} \vert z \vert ^{m+2}+z_{i} \vert z \vert ^{m+2}-\xi _{i} \vert z \vert ^{m+2}}{ \vert z \vert ^{m+2} \vert \xi \vert ^{m+2}} \biggr\vert \\ &\quad\leq\frac{ |z_{i}|| |\xi|^{m+2}- |z|^{m+2}|+|z_{i}-\xi _{i}| |z|^{m+2}}{ |z|^{m+2} |\xi|^{m+2}} \\ &\quad\leq\frac{|z|||\xi|-|z||(|\xi|^{m+1}+|\xi |^{m}|z|+\cdots+|z|^{m+1})+|z-\xi||z||\xi||z|^{m}}{|z|^{m+2}|\xi |^{m+2}} \\ &\quad\leq \biggl\vert \frac{|z|||\xi|-|z||(|\xi|^{m+1}+|\xi |^{m}|z|+\cdots+|\xi||z|^{m})+|z-\xi||z||\xi||z|^{m}}{|z|^{m+2}|\xi |^{m+2}} \biggr\vert \\ &\quad\leq \biggl\vert \frac{|z-\xi| [(|\xi|^{m}+|\xi|^{m-1}|z|+\cdots +|z|^{m})+|z|^{m} ]}{|z|^{m+1}|\xi|^{m+1}} \biggr\vert \\ &\quad\leq \biggl\vert \frac{|z-\xi| [P_{m}(z,\xi)+|z|^{\frac{m}{2}}|\xi |^{\frac{m}{2}} ]}{|z|^{m+1}|\xi|^{m+1}} \biggr\vert . \end{aligned}$$
When \(|\xi|\leq|z|\), insert \(\xi_{i}|\xi|^{m+2}\) in the above formula, we can prove the above inequality in a similar way. □
Remark 1
Because the original Hile lemma cannot be used directly in the complex Clifford analysis, we give the conclusion of Theorem 3.2 which is similar to the classical Hile lemma and plays an important role in proving the properties of T-operators and Cauchy operators. We insert the appropriate items according to the situation and prove that inequality (2) holds. Inequality (2) is similar to the Hile lemma of the classical real vector and is complete symmetry with respect to the variable ξ, z. It is a good tool to prove the Hölder continuity of the T operator with B-M kernel in the complex Clifford analysis.
Theorem 3.3
Let
\(\Omega\subset C^{n+1}\)
be a bounded domain, \(\varphi\in L^{p}(\Omega)\), \(p>2(n+1)\), then for any
\(z_{1},z_{2}\in\Omega\), we have
and
Tφ
is Hölder continuous on Ω, where
\(\alpha= 1-\frac{2(n+1)}{p}\).
$$\begin{aligned} \bigl\vert (T\varphi) (z_{1})-(T\varphi) (z_{2}) \bigr\vert \leq M_{2} \Vert \varphi \Vert _{\Omega,p} \vert z_{1}-z_{2} \vert ^{\alpha}, \end{aligned}$$
(3)
Proof
Case 1. When \(|z_{1}-z_{2}|\geq1\), using Theorem 3.2 we have
$$\begin{aligned} \bigl\vert T\varphi(z_{1})-T\varphi(z_{2}) \bigr\vert \leq& 2M_{1} \Vert \varphi \Vert _{\Omega,p} \\ \leq& 2M_{1} \Vert \varphi \Vert _{\Omega,p} \frac {1}{ \vert z_{1}-z_{2} \vert ^{\alpha}} \vert z_{1}-z_{2} \vert ^{\alpha} \\ \leq& M_{2} \Vert \varphi \Vert _{\Omega,p} \vert z_{1}-z_{2} \vert ^{\alpha}. \end{aligned}$$
Case 2. When \(|z_{1}-z_{2}|<1\), we have
Let
$$\begin{aligned} & \bigl\vert T\varphi(z_{1})-T\varphi(z_{2}) \bigr\vert \\ &\quad\leq\frac{1}{\omega_{2n+2}2^{n+1}} \int_{\Omega} \bigl\vert \varphi(\xi ) \bigr\vert \biggl\vert \frac{\sum_{k=0}^{n}(\xi_{k}-z_{1k})\bar{e}_{k}}{ \vert \xi -z_{1} \vert ^{2n+2}} -\frac{\sum_{k=0}^{n}({\xi_{k}-z_{2k}})\bar{e}_{k}}{ \vert \xi -z_{2} \vert ^{2n+2}} \biggr\vert \vert d\bar{\xi}\wedge d{ \xi} \vert \\ &\qquad{}+ \frac{1}{\omega_{2n+2}2^{n+1}} \int_{\Omega} \bigl\vert \varphi(\xi ) \bigr\vert \biggl\vert \frac{\sum_{k=0}^{n}(\overline{\xi_{k}-z_{1k}})\bar{e}_{k}}{ \vert \xi -z_{1} \vert ^{2n+2}} -\frac{\sum_{k=0}^{n}(\overline{\xi_{k}-z_{2k}})\bar {e}_{k}}{ \vert \xi-z_{2} \vert ^{2n+2}} \biggr\vert \vert d\bar{\xi}\wedge d{ \xi} \vert \\ &\quad\leq\frac{1}{\omega_{2n+2}}\sum_{k=0}^{n} \int_{\Omega} \bigl\vert \varphi (\xi) \bigr\vert \biggl\vert \frac{(\xi_{k}-z_{1k})}{ \vert \xi-z_{1} \vert ^{2n+2}} -\frac{({\xi _{k}-z_{2k}})}{ \vert \xi-z_{2} \vert ^{2n+2}} \biggr\vert \,dV_{\xi} \\ &\qquad{}+\frac{1}{\omega_{2n+2}}\sum_{k=0}^{n} \int_{\Omega} \bigl\vert \varphi (\xi) \bigr\vert \biggl\vert \frac{(\overline{\xi_{k}-z_{1k}})}{ \vert \xi-z_{1} \vert ^{2n+2}} -\frac{(\overline{\xi_{k}-z_{2k}})}{ \vert \xi-z_{2} \vert ^{2n+2}} \biggr\vert \,dV_{\xi} \\ &\qquad=\frac{2}{\omega_{2n+2}}\sum_{k=0}^{n} \int_{\Omega} \bigl\vert \varphi(\xi ) \bigr\vert \biggl\vert \frac{(\xi_{k}-z_{1k})}{ \vert \xi-z_{1} \vert ^{2n+2}} -\frac{({\xi _{k}-z_{2k}})}{ \vert \xi-z_{2} \vert ^{2n+2}} \biggr\vert \,dV_{\xi}. \end{aligned}$$
$$I= \int_{\Omega} \bigl\vert \varphi(\xi) \bigr\vert \biggl\vert \frac{(\xi_{k}-z_{1k})}{ \vert \xi -z_{1} \vert ^{2n+2}} -\frac{({\xi_{k}-z_{2k}})}{ \vert \xi-z_{2} \vert ^{2n+2}} \biggr\vert \,dV_{\xi}. $$
According to Theorem 3.2, we can get
For \(I_{1}\), we have
Using Hölder’s inequality we have
Because \(p>2n+2\), \(\frac{1}{p}+\frac{1}{q}=1\), we can get
So
\(0\leq k\leq2n\), we get
and
By Hadamard’s lemma, we have
So we have
As to \(I_{2}\), using Hölder’s inequality we have
Since \(p>2n+2\) and \(\frac{1}{p}+\frac{1}{q}=1\), we get
So
From Hadamard’s lemma, we get
So
Hence
Using Hölder’s inequality, we obtain
□
$$\begin{aligned} I \leq& \int_{\Omega} \bigl\vert \varphi(\xi) \bigr\vert \biggl\vert \frac{ \vert z_{1}-z_{2} \vert [P_{2n}(\xi-z_{1},\xi-z_{2})+ \vert \xi-z_{1} \vert ^{n} \vert \xi-z_{2} \vert ^{n} ]}{ \vert \xi -z_{1} \vert ^{2n+1} \vert \xi-z_{2} \vert ^{2n+1}} \biggr\vert \,dV_{\xi} \\ =& \vert z_{1}-z_{2} \vert \int_{\Omega} \bigl\vert \varphi(\xi) \bigr\vert \biggl\vert \frac{P_{2n}(\xi -z_{1},\xi-z_{2})}{ \vert \xi-z_{1} \vert ^{2n+1} \vert \xi-z_{2} \vert ^{2n+1}} \biggr\vert \,dV_{\xi } \\ &{}+ \vert z_{1}-z_{2} \vert \int_{\Omega} \bigl\vert \varphi(\xi) \bigr\vert \frac{ \vert \xi-z_{1} \vert ^{n} \vert \xi -z_{2} \vert ^{n}}{ \vert \xi-z_{1} \vert ^{2n+1} \vert \xi-z_{2} \vert ^{2n+1}}\,dV_{\xi} \\ =&I_{1}+I_{2}. \end{aligned}$$
$$\begin{aligned} I_{1} =& \vert z_{1}-z_{2} \vert \int_{\Omega}\sum_{k=0}^{2n} \vert \xi-z_{1} \vert ^{-(k+1)}|\xi-z_{2})|^{-(2n+1-k)}| \varphi(\xi)|dV_{\xi} \\ =& \vert z_{1}-z_{2} \vert \sum _{k=0}^{2n} \int_{\Omega} \vert \xi-z_{1} \vert ^{-(k+1)}| \xi-z_{2})|^{-(2n+1-k)}|\varphi(\xi)|dV_{\xi}. \end{aligned}$$
$$I_{1}\leq \vert z_{1}-z_{2} \vert \Vert \varphi \Vert _{\Omega,p} \sum_{k=0}^{2n} \biggl( \int_{\Omega} \vert \xi-z_{1} \vert ^{-(k+1)q} \vert \xi -z_{2} \vert ^{-(2n+1-k)q} \,dV_{\xi} \biggr)^{\frac{1}{q}}. $$
$$1< q< \frac{2n+2}{2n+1}. $$
$$2n+1< (2n+1)q< 2n+2, $$
$$\begin{gathered} (k+1)q\leq2n+2, \\ (2n+1-k)q\leq2n+2, \end{gathered} $$
$$(k+1)q+(2n+1-k)q>2n+2. $$
$$\begin{aligned} & \int_{\Omega} \vert \xi-z_{1} \vert ^{-(k+1)q}| \xi-z_{2})|^{-(2n+1-k)q}\,dV_{\xi} \\ &\quad\leq K_{6} \vert z_{1}-z_{2} \vert ^{(2n+2)-(2n+1-k)q-(k+1)q} \\ &\quad=K_{6} \vert z_{1}-z_{2} \vert ^{(2n+2)-(2n+2)q}. \end{aligned}$$
$$\begin{aligned} I_{1} \leq& (2n+1)K_{6} \Vert \varphi \Vert _{\Omega,p} \vert z_{1}-z_{2} \vert ^{1+\frac{2n+2}{q}-(2n+2)} \\ \leq&(2n+1)K_{6} \Vert \varphi \Vert _{\Omega,p} \vert z_{1}-z_{2} \vert ^{1-\frac{2n+2}{p}}. \end{aligned}$$
$$\begin{aligned} I_{2} =& \vert z_{1}-z_{2} \vert \int_{\Omega} \bigl\vert \varphi(\xi) \bigr\vert \frac{ \vert \xi -z_{1} \vert ^{n} \vert \xi-z_{2} \vert ^{n}}{ \vert \xi-z_{1} \vert ^{2n+1} \vert \xi-z_{2} \vert ^{2n+1}}\,dV_{\xi } \\ \leq& \vert z_{1}-z_{2} \vert \Vert \varphi \Vert _{\Omega,p} \biggl( \int_{\Omega } \vert \xi-z_{1} \vert ^{-(n+1)q} \vert \xi-z_{2} \vert ^{-(n+1)q} \,dV_{\xi} \biggr)^{\frac{1}{q}}. \end{aligned}$$
$$1< q< \frac{2n+2}{2n+1}. $$
$$\begin{gathered} (n+1)q\leq2n+2, \\ (2n+2)q\geq2n+2. \end{gathered} $$
$$\begin{aligned} & \int_{\Omega} \vert \xi-z_{1} \vert ^{-(n+1)q} \vert \xi-z_{2} \vert ^{-(n+1)q} \,dV_{\xi} \\ &\quad\leq K_{7} \vert z_{1}-z_{2} \vert ^{(2n+2)-(2n+2)q}. \end{aligned}$$
$$\begin{aligned} I_{2} \leq& K_{7} \Vert \varphi \Vert _{\Omega,p} \vert z_{1}-z_{2} \vert ^{1+\frac{2n+2}{q}-(2n+2)} \\ \leq& K_{7} \Vert \varphi \Vert _{\Omega,p} \vert z_{1}-z_{2} \vert ^{1-\frac{2n+2}{p}}. \end{aligned}$$
$$\begin{aligned} I =&I_{1}+I_{2} \\ \leq& (2n+1) (K_{6}+K_{7}) \Vert \varphi \Vert _{\Omega,p} \vert z_{1}-z_{2} \vert ^{1-\frac{2n+2}{p}} \\ =&K_{8} \Vert \varphi \Vert _{\Omega,p} \vert z_{1}-z_{2} \vert ^{1-\frac {2n+2}{p}}. \end{aligned}$$
$$\begin{aligned} \bigl\vert T\varphi(z_{1})-T\varphi(z_{2}) \bigr\vert \leq&\frac{2}{\omega_{2n+2}} K_{8} \Vert \varphi \Vert _{\Omega,p} \vert z_{1}-z_{2} \vert ^{1-\frac{2n+2}{p}} \\ \leq&K_{9} \Vert \varphi \Vert _{\Omega,p} \vert z_{1}-z_{2} \vert ^{1-\frac{2n+2}{p}} \\ \leq&M_{2} \Vert \varphi \Vert _{\Omega,p} \vert z_{1}-z_{2} \vert ^{\alpha}. \end{aligned}$$
Remark 2
In Case 2 of this theorem, we use the inequality of Theorem 3.3, Hölder’s inequality, and Hadamard’s lemma. This result enriches the theoretical system of the complex Clifford analysis.
Theorem 3.4
Let
\(\Omega\subset C^{n+1}\)
be a bounded domain, \(\varphi\in L^{p}(\Omega)\), \(1< p<2n+2\), γ
is an arbitrary constant which satisfies
\(1<\gamma<\frac{(2n+2)p}{(2n+2)-p}\), then
Tφ
is
γ-integrable on Ω, that is, \(T\varphi \in L^{\gamma}(\Omega)\), and the following inequality
is true.
$$\begin{aligned} \Vert T\varphi \Vert _{\Omega,\gamma}\leq M_{3} \Vert \varphi \Vert _{\Omega,p} \end{aligned}$$
(4)
Proof
For convenience, we introduce the notation b, suppose \(b=\frac {1}{\gamma}-\frac{1}{p}+\frac{1}{2n+2}\), then from\(1<\gamma<\frac{(2n+2)p}{(2n+2)-p}\) we know \(b>0\). Here are two cases to prove that Tφ is γ-integrable on Ω.
Case 1. When \(p<\gamma<\frac{(2n+2)p}{(2n+2)-p}\), \(0<\frac{p}{\gamma }<1\), thus \(0< p(\frac{1}{p}-\frac{1}{\gamma})=1-\frac{p}{\gamma}<1\), again
Choose \(q>0\) such that \(\frac{1}{p}+\frac{1}{q}=1\), then we have
Therefore,
Because \(1< p<\gamma\), \(\frac{1}{\gamma}+(\frac{1}{p}-\frac{1}{\gamma })+\frac{1}{q}=1\), using Hölder’s inequality we get
Because \(b>0\), we have
From Lemma 2.2 we can know that two integrals are meaningful, we assume that \(K_{11}= \sup_{\xi\in\Omega} \int_{\Omega}|\xi -z|^{(2n+2)(\frac{qb}{2}-1)}\,dV_{\xi}\).
$$\frac{p}{\gamma}+p\biggl(\frac{1}{p}-\frac{1}{\gamma}\biggr)=1. $$
$$\begin{aligned} &(2n+2) \biggl(\frac{b}{2}-\frac{1}{\gamma}\biggr)+(2n+2) \biggl( \frac{b}{2}-\frac{1}{q}\biggr) \\ &\quad=(2n+2) \biggl(b-\frac{1}{\gamma}-\frac{1}{q}\biggr) \\ &\quad=(2n+2) \biggl(\frac{1}{\gamma}-\frac{1}{p}+\frac{1}{2n+2}- \frac {1}{\gamma}-\frac{1}{q}\biggr) \\ &\quad=(2n+2) \biggl(-1+\frac{1}{2n+2}\biggr) \\ &\quad=-(2n+1). \end{aligned}$$
$$\begin{aligned} & \bigl\vert T\varphi(z) \bigr\vert \\ &\quad= \frac{1}{2^{n+1}\omega_{2n+2}} \biggl\vert \int_{\Omega}\varphi(\xi ) \biggl(\frac{\sum_{k=0}^{n}(\xi_{k}-z_{k})\bar{e}_{k}}{ \vert \xi-z \vert ^{2n+2}}+ \frac{\sum_{k=0}^{n}(\overline{\xi_{k}-z_{k}})\bar{e}_{k}}{ \vert \xi -z \vert ^{2n+2}} \biggr)\,d\bar{\xi}\wedge d{\xi} \biggr\vert \\ &\quad\leq\frac{K_{10}}{\omega_{2n+2}} \biggl( \int_{\Omega} \bigl\vert \varphi (\xi) \bigr\vert \frac{1}{ \vert \xi-z \vert ^{2n+1}}\,dV_{\xi} + \int_{\Omega} \bigl\vert \varphi (\xi) \bigr\vert \frac{1}{ \vert \xi-z \vert ^{2n+1}}\,dV_{\xi} \biggr) \\ &\quad= \frac{2K_{10}}{\omega_{2n+2}} \int_{\Omega} \bigl\vert \varphi(\xi ) \bigr\vert \frac{1}{ \vert \xi-z \vert ^{2n+1}}\,dV_{\xi} \\ &\quad= \frac{2K_{10}}{\omega_{2n+2}} \int_{\Omega} \bigl\vert \varphi(\xi ) \bigr\vert ^{\frac{p}{\gamma}} \vert \xi-z \vert ^{(2n+2)(\frac{b}{2}-\frac{1}{\gamma})} \bigl\vert \varphi(\xi) \bigr\vert ^{p(\frac{1}{p}-\frac{1}{\gamma})} \vert \xi -z \vert ^{(2n+2)(\frac{b}{2}-\frac{1}{q})}\,dV_{\xi}. \end{aligned}$$
$$\begin{aligned} & \bigl\vert T\varphi(z) \bigr\vert \\ &\quad\leq\frac{2K_{10}}{\omega_{2n+2}} \biggl( \int_{\Omega} \bigl\vert \varphi(\xi) \bigr\vert ^{p} \vert \xi-z \vert ^{(2n+2)(\frac{\gamma b}{2}-1)}\,dV_{\xi } \biggr)^{\frac{1}{\gamma}} \biggl( \int_{\Omega} \bigl\vert \varphi(\xi) \bigr\vert ^{p}\,dV_{\xi} \biggr)^{\frac {1}{p}-\frac{1}{\gamma}} \\ & \qquad{} \cdot \biggl( \int_{\Omega} \vert \xi-z \vert ^{(2n+2)(\frac {qb}{2}-1)}\,dV_{\xi} \biggr)^{\frac{1}{q}} \\ &\quad= \frac{2K_{10}}{\omega_{2n+2}} \biggl( \int_{\Omega} \bigl\vert \varphi (\xi) \bigr\vert ^{p} \vert \xi-z \vert ^{(2n+2)(\frac{\gamma b}{2}-1)}\,dV_{\xi} \biggr)^{\frac{1}{\gamma}} \Vert \varphi \Vert _{\Omega,p}^{1-\frac{p}{\gamma }} \\ & \qquad{} \cdot \biggl( \int_{\Omega} \vert \xi-z \vert ^{(2n+2)(\frac {qb}{2}-1)}\,dV_{\xi} \biggr)^{\frac{1}{q}}. \end{aligned}$$
$$\begin{gathered} (2n+2) \biggl(1-\frac{\gamma b}{2}\biggr)< 2n+2, \\ (2n+2) \biggl(1-\frac{qb}{2}\biggr)< 2n+2. \end{gathered} $$
Therefore we have
Let \(K_{12}= \sup_{\xi\in\Omega} \int_{\Omega}|\xi -z|^{(2n+2)(\frac{\gamma b}{2}-1)}\,dV_{z}\), so we have
where \(K_{13}= (\frac{2}{\omega_{2n+2}} )^{\gamma}K_{11}^{\frac {\gamma}{q}}K_{12}\).
$$\bigl\vert T\varphi(z) \bigr\vert ^{\gamma}\leq \biggl( \frac{2}{\omega_{2n+2}} \biggr)^{\gamma }K_{11}^{\frac{\gamma}{q}} \Vert \varphi \Vert _{\Omega,p}^{\gamma-p} \biggl( \int_{\Omega} \bigl\vert \varphi(\xi) \bigr\vert ^{p} \vert \xi-z \vert ^{(2n+2)(\frac{\gamma b}{2}-1)}\,dV_{\xi } \biggr). $$
$$\bigl\vert T\varphi(z) \bigr\vert ^{\gamma}\leq K_{12} \Vert \varphi \Vert _{\Omega,p}^{\gamma -p} \Vert \varphi \Vert _{\Omega,p}^{p}=K_{13} \Vert \varphi \Vert _{\Omega,p}^{\gamma}, $$
Hence, we get
where \(K_{14}= K_{13}^{\gamma}\).
$$\Vert T\varphi \Vert _{\Omega,\gamma}= \biggl( \int_{\Omega} \bigl\vert T\varphi (z) \bigr\vert ^{\gamma}\,dV_{\xi} \biggr)^{\frac{1}{\gamma}} \leq K_{13}^{\gamma} \Vert \varphi \Vert _{\Omega,p}=K_{14} \Vert \varphi \Vert _{\Omega,p}, $$
(2) When \(p\geq\gamma>1\), choose m such that \(0<\frac{(2n+2)\gamma }{(2n+2)+\gamma}<m<\gamma\), and m is an arbitrary positive constant satisfying \(m<\gamma<\frac {(2n+2)m}{(2n+2)+m}\). Because \(\varphi\in L^{p}(\Omega)\), \(m< p\), we have \(\varphi\in L^{m}(\Omega)\).
Choose \(\frac{1}{p}+\frac{1}{q}=\frac{1}{m}\). Therefore, from the proof process of (1) and Lemma 2.5, we get
Therefore Tφ is γ integrable on Ω. If we choose \(M_{3}=\max\{K_{14},K_{16}\}\), then
□
$$\begin{aligned} & \Vert T\varphi \Vert _{\Omega,p} \\ &\quad\leq K_{15} \Vert \varphi \Vert _{\Omega,m} \\ &\quad= K_{15} \biggl[ \int_{\Omega} \bigl\vert \varphi(\xi)\cdot1 \bigr\vert ^{m} \biggr]^{\frac{1}{m}} \\ &\quad\leq K_{15} \biggl[ \int_{\Omega} \bigl\vert \varphi(\xi)\cdot1 \bigr\vert ^{p} \biggr]^{\frac{1}{p}} \biggl[ \int_{\Omega} \bigl\vert \varphi(\xi)\cdot1 \bigr\vert ^{q} \biggr]^{\frac{1}{q}} \\ &\quad\leq K_{15}V_{\Omega}^{\frac{1}{q}} \Vert \varphi \Vert _{\Omega,p} \\ &\quad\leq K_{16} \Vert \varphi \Vert _{\Omega,p}. \end{aligned}$$
$$\Vert T\varphi \Vert _{\Omega,\gamma}\leq M_{3} \Vert \varphi \Vert _{\Omega,p}. $$
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