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1. 信阳师范学院数学与统计学院, 河南 信阳 464000;
2. 中国科学院大学数学科学学院, 北京 100049
摘要: 证明,对于任意一个非零
关键词: Riesz变换$\mathscr{D}({{\mathbb{R}}^{n}})$紧支集
For f∈Lp(
$Hf\left( x \right): = \frac{1}{{\rm{ \mathsf{ π} }}}{\rm{p}}.{\rm{v}}.\int_\mathbb{R} {\frac{{f\left( t \right)}}{{x - t}}{\rm{d}}t} ,$ | (1) |
${\rm{p}}.{\rm{v}}.\int_\mathbb{R} {\frac{{f\left( t \right)}}{{x - t}}{\rm{d}}t} = \mathop {\lim }\limits_{\varepsilon \to 0} \int_{\left| {x - t} \right| > \varepsilon } {\frac{{f\left( t \right)}}{{x - t}}{\rm{d}}t} .$ |
$\begin{array}{*{20}{c}} {\mathscr{D}\left( {{\mathbb{R}^n}} \right) = \left\{ {\phi :\phi \in C_c^\infty \left( {{\mathbb{R}^n}} \right),} \right.} \\ {\left. {\forall \alpha \in N_0^n,{\rho _\alpha }\left( \phi \right) = \mathop {\sup }\limits_{x \in {\mathbb{R}^n}} \left| {{D^\alpha }\phi \left( x \right)} \right| < \infty } \right\}} \end{array}$ |
That is to say the function Hf does not have compact support for all non-zero function f∈
Consider n-dimensional Euclidean space
${H_n}f\left( x \right): = \frac{1}{{{{\rm{ \mathsf{ π} }}^n}}}{\rm{p}}.{\rm{v}}.\int_\mathbb{R} {\frac{{f\left( t \right)}}{{\left( {{x_1} - {t_1}} \right)\left( {{x_2} - {t_2}} \right) \cdots \left( {{x_n} - {t_n}} \right)}}{\rm{d}}t} .$ | (2) |
$\mathscr{D}\left( {{\mathbb{R}^2}} \right) \cap {H_2}\left( {\mathscr{D}\left( {{\mathbb{R}^2}} \right)} \right) = \left\{ 0 \right\}.$ |
$\mathscr{D}\left( {{\mathbb{R}^n}} \right) \cap {H_n}\left( {\mathscr{D}\left( {{\mathbb{R}^n}} \right)} \right) = \left\{ 0 \right\}.$ | (3) |
$\widehat {Hf}\left( \xi \right) = - i \cdot {\mathop{\rm sgn}} \left( \xi \right)\hat f\left( \xi \right),$ | (4) |
$\rm{sgn}\left( x \right) = \left\{ \begin{array}{l}1\;\;\;\;\;\;\;{\rm{if}}\;\;x > 1,\\0\;\;\;\;\;\;\;{\rm{if}}\;\;n = 0,\\ - 1\;\;\;\;\;{\rm{if}}\;\;x < 0.\end{array} \right.$ |
${R_j}f\left( x \right): = \frac{{\Gamma \left( {\frac{{n + 1}}{2}} \right)}}{{{{\rm{ \mathsf{ π} }}^{\frac{{n + 1}}{2}}}}}{\rm{p}}.{\rm{v}}.\int_\mathbb{R} {\frac{{{x_j} - {y_j}}}{{{{\left| {x - y} \right|}^{n + 1}}}}f\left( y \right){\rm{d}}y} ,$ |
$\widehat {{R_j}f}\left( \xi \right) = - i \cdot \frac{{{\xi _j}}}{{\left| \xi \right|}}\hat f\left( \xi \right).$ | (5) |
$\mathscr{D}\left( {{\mathbb{R}^n}} \right) \cap {R_j}\left( {\mathscr{D}\left( {{\mathbb{R}^n}} \right)} \right) = \left\{ 0 \right\}$ | (6) |
In section 1, we will give an affirmative answer for (6).
1 Main resultsBefore we prove our main theorem, we need the following lemmas.
Lemma 1.1?? Suppose that f∈L1(
This lemma is a basic result which appears in many books on real analysis and we refer readers to Ref.[5].
Lemma 1.2?? Suppose that f, g∈C∞(
Proof?? Considering the continuity of the function g, we have
$g\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \frac{{{x_j}}}{{\left| x \right|}}f\left( 0 \right).$ | (7) |
When |α|=1, taking derivatives on g, we have
$\begin{array}{*{20}{c}}{\frac{{\partial g}}{{\partial {x_i}}}\left( x \right) = \frac{{\partial f}}{{\partial {x_i}}}\left( x \right)\frac{{{x_j}}}{{\left| x \right|}} + f\left( x \right)\frac{\partial }{{\partial {x_i}}}\left( {\frac{{{x_j}}}{{\left| x \right|}}} \right),i \ne j;}\\{\frac{{\partial g}}{{\partial {x_j}}}\left( x \right) = \frac{{\partial f}}{{\partial {x_j}}}\left( x \right)\frac{{{x_j}}}{{\left| x \right|}} + f\left( x \right)\frac{\partial }{{\partial {x_j}}}\left( {\frac{{{x_j}}}{{\left| x \right|}}} \right).}\end{array}$ | (8) |
$\begin{array}{*{20}{c}}{\frac{{\partial g}}{{\partial {x_i}}}\left( 0 \right) = \frac{{\partial f}}{{\partial {x_i}}}\left( 0 \right)\mathop {\lim }\limits_{x \to 0} \frac{{{x_j}}}{{\left| x \right|}},i \ne j;}\\{\frac{{\partial g}}{{\partial {x_j}}}\left( 0 \right) = \frac{{\partial f}}{{\partial {x_j}}}\left( 0 \right)\mathop {\lim }\limits_{x \to 0} \frac{{{x_j}}}{{\left| x \right|}}.}\end{array}$ | (9) |
$\frac{{\partial f}}{{\partial {x_i}}}\left( 0 \right) = 0,$ | (10) |
Now we use induction method to solve the case |α|≥2.
Suppose that for any multiple-index β with |β| < |α|, there holds
${\left( {\frac{\partial }{{\partial x}}} \right)^\beta }f\left( 0 \right) = 0.$ | (11) |
$\begin{array}{*{20}{c}}{{{\left( {\frac{\partial }{{\partial x}}} \right)}^\alpha }g\left( x \right) = \frac{{{x_j}}}{{\left| x \right|}}{{\left( {\frac{\partial }{{\partial x}}} \right)}^\alpha }f\left( x \right) + }\\{\sum\limits_{\gamma + \delta = \alpha } {{{\left( {\frac{\partial }{{\partial x}}} \right)}^\gamma }f\left( x \right){{\left( {\frac{\partial }{{\partial x}}} \right)}^\delta }\left( {\frac{{{x_j}}}{{\left| x \right|}}} \right)} .}\end{array}$ | (12) |
${\left( {\frac{\partial }{{\partial x}}} \right)^\alpha }g\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \frac{{{x_j}}}{{\left| x \right|}}{\left( {\frac{\partial }{{\partial x}}} \right)^\alpha }f\left( 0 \right).$ | (13) |
Now we shall prove our main theorem.
Theorem 1.1?? Suppose that f∈
Proof ??First we obtain f∈L2(
Since the Riesz transform Rj is of strong type (p, p) with 1 < p < ∞, we obtain that Rjf is also in L2(
Since f∈
Noting
${\left( {\frac{\partial }{{\partial x}}} \right)^\alpha }\hat f\left( 0 \right) = 0,$ | (14) |
That is to say,
$\int_{{\mathbb{R}^n}} {{{\left( { - 2{\rm{ \mathsf{ π} }}ix} \right)}^\alpha }f\left( x \right){\rm{d}}x} = 0$ | (15) |
Let supp f∈Q, then the identity
$\int_Q {P\left( x \right)f\left( x \right){\rm{d}}x} = 0$ | (16) |
The collection of all continuous functions defined on Q is denoted by C(Q). Considering that the polynomials are dense in the functional space C(Q), it immediately follows f≡0 from (16).
Obviously, Theorem 1.1 implies (6), which is our main conclusion.
References
[1] | Grafakos L. Classical and modern Fourier analysis[M].New Jersey: Pearson Education Inc, 2004. |
[2] | Yang L. A distribution space for Hilbert transform and its applications[J].Science in China Series A:Mathematics, 2008, 51(12):2217–2230.DOI:10.1007/s11425-008-0007-1 |
[3] | Cui X, Wang R, Yan D. Some properties for double Hilbert transform on |
[4] | Shen F. High-dimensional Hilbert transform on |
[5] | Royden H L, Fitzpatrick P. Real analysis[M].New York: Macmillan, 1988. |