For f∈L^p($\mathbb{R}$), 1≤p < ∞, the classical Hilbert transform is defined as

$ 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)

where "p.v."is the Cauchy principal value (see Ref.[1]), that is

$ {\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} . $

Yang^[2] proved that $\mathscr{D}$ ($\mathbb{R}$)∩H($\mathscr{D}$($\mathbb{R}$))={0}, where

$ \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} $

for n∈N.

That is to say the function Hf does not have compact support for all non-zero function f∈$\mathscr{D}$($\mathbb{R}$).

Consider n-dimensional Euclidean space $\mathbb{R}$ⁿ (n∈N). For f∈L^p($\mathbb{R}$ⁿ), 1≤p < ∞, x=(x₁, x₂, …, x_n)∈$\mathbb{R}$ⁿ, t=(t₁, t₂, …, t_n)∈$\mathbb{R}$ⁿ, the n-dimensional Hilbert transform H_n is defined as

$ {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)

For n=2 Cui et al.^[3] obtained the result:

$ \mathscr{D}\left( {{\mathbb{R}^2}} \right) \cap {H_2}\left( {\mathscr{D}\left( {{\mathbb{R}^2}} \right)} \right) = \left\{ 0 \right\}. $

When n>2, Shen^[4] proved

$ \mathscr{D}\left( {{\mathbb{R}^n}} \right) \cap {H_n}\left( {\mathscr{D}\left( {{\mathbb{R}^n}} \right)} \right) = \left\{ 0 \right\}. $

(3)

We have known that Hilbert transform has a relation with Fourier transform which can be presented as

$ \widehat {Hf}\left( \xi \right) = - i \cdot {\mathop{\rm sgn}} \left( \xi \right)\hat f\left( \xi \right), $

(4)

where

$ \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. $

For f∈L^p($\mathbb{R}$ⁿ), 1≤p < ∞, x=(x₁, x₂, …, x_n)∈$\mathbb{R}$ⁿ, the Riesz transform R_j (j∈{1, 2, …, n}) is defined as

$ {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} , $

and we have (see Ref.[1])

$ \widehat {{R_j}f}\left( \xi \right) = - i \cdot \frac{{{\xi _j}}}{{\left| \xi \right|}}\hat f\left( \xi \right). $

(5)

If n=1, the identity (5) will reduce to (4). In this sense, Riesz transform is the higher-dimensional form of Hilbert transform. Motivated by Refs.[2-4], we shall consider whether the equation

$ \mathscr{D}\left( {{\mathbb{R}^n}} \right) \cap {R_j}\left( {\mathscr{D}\left( {{\mathbb{R}^n}} \right)} \right) = \left\{ 0 \right\} $

(6)

holds for n>1.

In section 1, we will give an affirmative answer for (6).

1 Main results

Before we prove our main theorem, we need the following lemmas.

Lemma 1.1   Suppose that f∈L¹($\mathbb{R}$ⁿ) with compact support. Then we have ${\hat{f}}$ ∈C^∞($\mathbb{R}$ⁿ).

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^∞($\mathbb{R}$ⁿ) and $g\left( x \right)=\frac{{{x}_{j}}}{|x|}f\left( x \right)$ for some j∈{1, …, n}. Then we have ${{\left( \frac{\partial }{\partial x} \right)}^{\alpha }}f\left( 0 \right)=0$ for all multi-index $\alpha \in \mathbb{N}_{0}^{n}$.

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)

The equality (7) implies that f(0)=0.

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)

By substituting x=0 into (8), there hold

$ \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)

Since $\frac{\partial g}{\partial {{x}_{i}}}\left( 0 \right)$ and $\frac{\partial g}{\partial {{x}_{j}}}\left( 0 \right)$ all exist and lim$\frac{{{x}_{i}}}{|x|}$, (i=1, …, n) does not exist while x tends to 0, we derive from (9) that

$ \frac{{\partial f}}{{\partial {x_i}}}\left( 0 \right) = 0, $

(10)

for all 1≤i≤n.

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)

Furthermore,

$ \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)

The assumption g∈C^∞($\mathbb{R}$ⁿ) implies that ${{\left( \frac{\partial }{\partial x} \right)}^{\alpha }}g\left( x \right)$ exists for all x∈$\mathbb{R}$ⁿ. Then we obtain from (11) and (12) that

$ {\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)

Thus we immediately get ${{\left( \frac{\partial }{\partial x} \right)}^{\alpha }}f\left( 0 \right)=0$ from (13).

Now we shall prove our main theorem.

Theorem 1.1   Suppose that f∈$\mathscr{D}$($\mathbb{R}$ⁿ). If R_jf has compact support, then $\widehat{{{R}_{j}}f}$ ∈C^∞($\mathbb{R}$ⁿ) and f≡0.

Proof   First we obtain f∈L²($\mathbb{R}$ⁿ). from f∈$\mathscr{D}$($\mathbb{R}$ⁿ).

Since the Riesz transform R_j is of strong type (p, p) with 1 < p < ∞, we obtain that R_jf is also in L²($\mathbb{R}$ⁿ). Furthermore, R_jf $\subset $ L¹($\mathbb{R}$ⁿ) from the assumption that R_jf has compact support. By Lemma 1.1, there holds $\widehat{{{R}_{j}}f}$ $\subset $ C^∞($\mathbb{R}$ⁿ).

Since f∈$\mathscr{D}$($\mathbb{R}$ⁿ) means that f∈L¹($\mathbb{R}$ⁿ) and f has compact support, by using Lemma 1.1 again, it yields ${\hat{f}}$∈C^∞($\mathbb{R}$ⁿ).

Noting $\widehat{{{R}_{j}}f}$ (ξ)= $-i\frac{{{\xi }_{j}}}{|\xi |}\hat{f}\left( \xi \right)$ by (5), we conclude

$ {\left( {\frac{\partial }{{\partial x}}} \right)^\alpha }\hat f\left( 0 \right) = 0, $

(14)

for all $\alpha \in \mathbb{N}_{0}^{n}$ by Lemma 1.2.

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)

holds for all $\alpha \in \mathbb{N}_{0}^{n}$.

Let supp f∈Q, then the identity

$ \int_Q {P\left( x \right)f\left( x \right){\rm{d}}x} = 0 $

(16)

holds for all polynomials P by (15).

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.