1. Introduction

The complex signal method especially analytic signal method is a classical way of defining the phase and amplitude of signals, which plays an important role in meteorological as well as atmospheric applications, ocean engineering, structural science, and imaging processing, one can refer to [16,17,18,22,28,24] for details. This results in the widely used empirical mode decomposition and the Hilbert-Huang transform, see for instance [7,16,17].

The Hilbert transform is a well-known and useful concept in harmonic analysis and signal processing (see for instance [13,2,9,15]). The 1-dimensional Hilbert transform $H$ for functions $f\in L^{p}(\mathbb{R})$ ($1\leq p ＜\infty$) can be stated as follows

$(Hf)(x):={\rm p.v.}\frac{1}{\pi}\int_{\mathbb{R}}\frac{f(y)}{x-y}dy=\lim_{\epsilon\rightarrow 0}\frac{1}{\pi}\int_{|y-x|\geq \epsilon}\frac{f(y)}{x-y}dy, \ x\in \mathbb{R}.$

Regarding to the Hilbert transform defined above: If $f, g\in L^{2}(\mathbb{R})$ satisfy either ${\rm{supp}} \hat{f}\subseteq\mathbb{R}_{+}(\mathbb{R}_{+}=[0, \infty))$, ${\rm{supp}} \hat{g}\subseteq\mathbb{R}_{+}$ or ${\rm{supp}}\hat{f}\subseteq[-a, a]$, ${\rm{supp}}\hat{g}\subseteq(-\infty, -a]\bigcup[a, \infty)$ for some positive number $a$, then the following identity holds true

$[H(fg)](x)=f(x)(Hg)(x), \ \ x\in\mathbb{R},$

which is named as the Bedrosian identity to honor Bedrosian for his contribution [1]. Later, some efforts were devoted to obtaining more general sufficient conditions (see e.g., [3,21]). The Bedrosian identity simplifies the calculation of the Hilbert transform of a product of functions. In recent years, the Bedrosian identity has attracted considerable attention and progress has been made. There is a large number of documents for the studies of the Bedrosian identity, see for example [6, 5, 8, 25, 26, 30, 31, 32, 34, 35, 36, 37].. It is worthwhile to state that an observation in [33] implies that the Hilbert transform is essentially the only operator satisfying the Bedrosian theorem.

It is well known that the complex signal method via the Hilbert transform has already become a significant tool in signal analysis and processing, especially in the time-frequency analysis (see, e.g., [4,1,9,14,21]). Imaging and other applications to multidimensional signals call for extension of the method to higher dimensions. Therefore, it is natural to establish the Bedrosian identity in n-dimension case. To the best knowledge of the authors, there are only very few results on multidimensional Bedrosian identities. Actually, in some special case, that is $p=2$, one can refer to [33,38,19] consulting the multidimensional Bedrosian identity.

Now let us give the definition of the total Hilbert transform as well as the partial Hilbert transform.

Definition 1.1 The partial Hilbert transform for $f\in L^{p}(\mathbb{R}^{n}) (1\leq p ＜\infty)$, $H_{j}f$ is given by

$(H_{j}f)(x):={\rm p.v.}\frac{1}{\pi}\int_{\mathbb{R}} \frac{f(y)}{(x_{j}-y_{j})}dy_{j}.$

The total Hilbert transform $H$ of a function is given by

$(Hf)(x):={\rm p.v.}\frac{1}{(\pi)^{n}}\int_{\mathbb{R}^{n}} \frac{f(y)}{\prod_{j=1}^{n}(x_{j}-y_{j})}dy \buildrel \Delta \over = (H_{1}H_{2}\cdots H_{n})f(x), \ x\in \mathbb{R}^{n}.$

The Fourier transform $\hat{f}$ of $f\in L^{1}(\mathbb{R}^{n})$ is defined as

$\hat{f}(x)=\int_{\mathbb{R}^{n}}f(t)e^{-ix\cdot t}dt, \ \ x\in \mathbb{R}^{n}.$

Next, let us give some basic notation. Let $\mathcal{D}(\mathbb{R}^{n})$ be the space of infinitely differentiable functions on $\mathbb{R}^{n}$ with compact support and $\mathcal{D}'(\mathbb{R}^{n})$ the space of distributions, namely, the dual of $\mathcal{D}(\mathbb{R}^{n})$. A distribution $T$ is said to vanish on an open subset $\Omega\subseteq\mathbb{R}^{n}$ as long as for each $\phi\in\mathcal{D}(\mathbb{R}^{n})$ with ${\rm{supp}}\phi\subseteq\Omega$, $＜T, \phi>=T(\phi)$ equals zero. The support of $T\in\mathcal{D}'(\mathbb{R}^{n})$, denoted by ${\rm{supp}}T$, is defined to be the complement of the largest open subset of $\mathbb{R}^{n}$ on which $T$ vanishes. This definition is consistent with the ordinary one when $T$ is a continuous function. Set

$D_{+}=\{x:x=(x_{1}, \cdots, x_{n})\in\mathbb{R}^{n}, sgn(-x)=\prod_{j=1}^{n} sgn(-x_{j})=1\},$

$D_{-}=\{x:x=(x_{1}, \cdots, x_{n})\in\mathbb{R}^{n}, sgn(-x)=\prod_{j=1}^{n} sgn(-x_{j})=-1\}$

and

$D_{0}=\{x:x=(x_{1}, \cdots, x_{n})\in\mathbb{R}^{n}, sgn(-x)=\prod_{j=1}^{n} sgn(-x_{j})=0\}.$

We denote by $\mathcal{D}_{D_{+}}(\mathbb{R}^{n})$, $\mathcal{D}_{D_{-}}(\mathbb{R}^{n})$ and $\mathcal{D}_{D_{0}}(\mathbb{R}^{n})$ the set of functions in $\mathcal{D}(\mathbb{R}^{n})$ that are supported on $D_{+}, D_{-}$ and $D_{0}$, respectively.

To apply the Fourier transform, we also introduce the Schwartz class $\mathcal{S}(\mathbb{R}^{n})$ and its dual $\mathcal{S}'(\mathbb{R}^{n})$, the space of temperate distributions. The Schwartz class $\mathcal{S}(\mathbb{R}^{n})$ consists of infinitely differentiable function $\varphi$ on $\mathbb{R}^{n}$ for all $\alpha, \beta\in \mathbb{Z}_{+}^{n}$ satisfies

$\sup_{x\in\mathbb{R}^{n}}|x^{\alpha}D^{\beta}\varphi(x)|＜\infty,$

where $\alpha=(\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}), \beta=(\beta_{1}, \beta_{2}, \cdots, \beta_{n}),$ $\alpha_{j}$ and $\beta_{j}$ are nonnegative integers. The Fourier transform $\hat{\varphi}$ is a linear homeomorphism from $\mathcal{S}(\mathbb{R}^{n})$ onto itself. Meanwhile, the following identity holds

$(H\varphi)^{\wedge}(x)=(-i){\rm{sgn}}(x)\hat\varphi(x), \ \ \forall \varphi\in \mathcal{S}(\mathbb{R}^{n}).$

The Fourier transform $\mathcal{F}:\mathcal{S}'(\mathbb{R}^{n})→ \mathcal{S}'(\mathbb{R}^{n})$ defined as

$\left\langle {\left. {\hat \psi , \varphi } \right\rangle } \right. = \left\langle {\left. {\psi , \hat \varphi } \right\rangle } \right., \;\;\;\;\forall \varphi \in {\cal S}({^n})$

is a linear isomorphism from $\mathcal{S}'(\mathbb{R}^{n})$ onto itself. For $\psi\in\mathcal{S}'(\mathbb{R}^{n})$, $\forall\varphi\in \mathcal{S}(\mathbb{R}^{n})$, it is easy to check that

$= \left\langle {\check{\psi} ,\tilde \varphi } \right\rangle$ = $= \left\langle {\psi ,\hat \varphi } \right\rangle = \left\langle {\hat \psi ,\varphi } \right\rangle .$

Therefore in the sense of distribution, we obtain

$\tilde{\check{\psi}}=\hat{\psi},$

where $\tilde{\varphi}(x)=\varphi(-x)$. $\check{\psi}$ is the inverse Fourier transform defined as

$\left\langle {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\smile$}} \over \psi } ,\varphi } \right\rangle = \left\langle {\psi ,\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\smile$}} \over \varphi } } \right\rangle .$

For the detail properties of $\mathcal{S}(\mathbb{R}^{n})$ and $\mathcal{S}'(\mathbb{R}^{n})$, see for example [27,11,14].

A function $f$ defined on $\mathbb{R}^{n}$ belongs to $\mathcal{D}_{L^{p}}(\mathbb{R}^{n}), \, 1＜p＜\infty$ if and only if

(1) $f\in C^{\infty}(\mathbb{R}^{n}),$

(2) $D^{k}f\in L^{p}(\mathbb{R}^{n}), \, k=0, 1, \cdots.$

In the sequel for $1＜p＜\infty$, we denote by $\mathcal{D}^{'}_{L^{p}}(\mathbb{R}^{n})$ the dual of $\mathcal{D}_{L^{p^{'}}}(\mathbb{R}^{n})$, where $\frac{1}{p}+\frac{1}{p'}=1.$

Our second main component of this paper is to consider the boundary values of holomorphic functions in distribution (see for instance [20]). More precisely, we show that the distribution $f\in \mathcal{D}'_{L^{p}}(\mathbb{R}^{n})$ can be represented by functions in the Hardy space on tube.

Let $B$ denote an open connected subset of $\mathbb{R}^{n}$. The tube is defined by $T_{B}\doteq\{x+iy:x\in\mathbb{R}^{n}, y\in B\}.$ The Hardy space [29] on tube $T_{B}$ is defined as

$H^{p}(T_{B}):=\{f\in H(T_{B}):\|f\|_{H^{p}}＜\infty\},$

where $\|f\|_{H^{p}}=\sup_{y\in B}\big(\int_{\mathbb{R}^{n}}|f(x+iy)|^{p}dx\big)^{\frac{1}{p}}$, and $H(T_{B})$ consists of all the holomorphic functions on $T_{B}$.

Definition 1.2 Let $f \in {{\cal D}'_{{L^p}}}({^n}), 1p\infty .$ The distributional differentiation and Hilbert transform of $f$ are defined as

$\left\langle {{D^k}f, \varphi } \right\rangle = \left\langle {f, {{( - 1)}^{|k|}}{D^k}\varphi } \right\rangle ,$

and

$Hf, \varphi = f, {( - 1)^n}H\varphi , \;\;\forall \varphi \in {{\cal D}_{Lp'}}({^n})$

respectively.

Here we want to mention that Pandey [23] proved that $D^{k}f\in\mathcal{D}_{L^{p}}^{'}(\mathbb{R}^{n}), \, \, Hf\in\mathcal{D}_{L^{p}}^{'}(\mathbb{R}^{n})$ and Hilbert transform $H$ defined on $L^{p}(\mathbb{R}^{n})\ (p>1)$ is isomorphism from $\mathcal{D}_{L^{p}}(\mathbb{R}^{n})$ onto itself.

The present paper is structured as follows. In section 2, we characterize the Bedrosian Identity of total Hilbert tranform, which consists of several lemmas. In section 3, we prove distribution $f\in \mathcal{D}'_{L^{p}}$ can be represented by functions in the Hardy space on tube.

2. The bedrosian identity for $L^{p}(\mathbb{R}^{n})$ function

This part is motivated by the need of defining multidimensional complex signals. We define the complex signal of $f\in L^{p}$ through total Hilbert transform $H$ as $f+iHf$. In this section we investigate the multidimensional Bedrosian identity $H(fg)=fHg$ for $f\in L^{p}(\mathbb{R}^{n}), \ g\in L^{q}(\mathbb{R}^{n})$ with $1 ＜p, \, q＜\infty$. In particular, several necessary and sufficient conditions to guarantee the Bedrosian identity to be valid are obtained.

Lemma 2.1 Let $f\in L^{p}(\mathbb{R}^{n})$ for $1＜p＜\infty$, then

$H(f\ast g)(x)=(Hf\ast g)(x)\ \ a.e.\ \ g\in L^{1}(\mathbb{R}^{n}).$

Proof. According to the properties of Hilbert transform and convolution, it is not difficult to show that $f\ast g, \ Hf\ast g$ and $H(f\ast g)$ all belong to $L^{p}(\mathbb{R}^{n})$. Thus both sides of the above equality are well defined. By Fubini's theorem, for all $h\in L^{p'}(\mathbb{R}^{n})$, we have

$\eqalign{ & \int_{{^n}} H (f * g)(\xi )h(\xi )d\xi = {( - 1)^n}\int_{{^n}} {(Hh)} (\xi )(f * g)(\xi )d\xi \cr & = {( - 1)^n}\int_{{^n}} g (u)\int_{{^n}} {(Hh)} (\xi )f(\xi - u)d\xi du \cr & = \int_{{^n}} g (u)\int_{{^n}} h (\xi )(Hf)(\xi - u)d\xi du \cr & = \int_{{^n}} h (\xi )(Hf * g)(\xi )d\xi \cr}$

The proof of the lemma is completed.

Lemma 2.2 Let $1\leq p\leq2, f\in L^{p}(\mathbb{R}^{n})$. If there exists $g\in L^{p}(\mathbb{R}^{n})$ such that

$(-i)^{n}sgn(x) \hat{f}(x)=\hat{g}(x),$

(1)

then $Hf=g a.e..$

Proof. As $f\in L^{p}(\mathbb{R}^{n})$ and $g\in L^{p}(\mathbb{R}^{n})$ satisfy $(1)$, we can choose a sequence of functions $\{\phi_{j}, j\in \mathbb{N}\}$ that are infinitely differentiable with compact support satisfying for each $f, g\in L^{p}(\mathbb{R}^{n})$

$\lim_{j\rightarrow\infty}\|\phi_{j}\ast f-f\|_{L^{p}(\mathbb{R}^{n})}=0, \, \, \, \, \, \lim_{j\rightarrow\infty}\|\phi_{j}\ast g-g\|_{L^{p}(\mathbb{R}^{n})}=0,$

where $\phi_{j}\ast f$ is convolution of $\phi_{j}$ and $f$ given by

$\phi_{j}\ast f(x)=\int_{\mathbb{R}^{n}}\phi_{j}(x-t)f(t)dt= \int_{\mathbb{R}^{n}}\phi_{j}(t)f(x-t)dt.$

It is clear that

$(\phi_{j}\ast f)^{\wedge}(x)=\hat{\phi_{j}}(x)\hat{f}(x).$

The above identity together with $(1)$ implies that

$(-i)^{n}sgn(x)(\phi_{j}\ast f)^{\wedge}(x)=\hat{\phi_{j}}(x)\hat{g}(x)\ \ a.e.\ x\in \mathbb{R}^{n}.$

Thus

$Hf\ast \phi_{j}=H(\phi_{j}\ast f)=\phi_{j}\ast g\quad$

(2)

According to the property of Hilbert transform

$\|H(\phi_{j}\ast f)-Hf\|_{L^{p}(\mathbb{R}^{n})}\leq A_{p}\|\phi_{j}\ast f-f\|_{L^{p}(\mathbb{R}^{n})}\rightarrow 0\ \ \ (j\rightarrow \infty).$

Therefore,

$\|Hf-g\|_{L^{p}(\mathbb{R}^{n})}\leq\|Hf-H(\phi_{j}\ast f)\|_{L^{p}(\mathbb{R}^{n})}+\|g\ast\phi_{j}-g\|_{L^{p}(\mathbb{R}^{n})}\rightarrow 0\ \ \ (j\rightarrow\infty).$

This fact yields that $Hf=g \ a.e$. We thus complete the proof.

Lemma 2.3 Assume that $p, q, r\in(1, \infty]$ satisfy $\frac{1}{p}+\frac{1}{q}=\frac{1}{r}\leq 1$. Let $f\in L^{p}(\mathbb{R}^{n})$ and $g\in L^{q}(\mathbb{R}^{n})$, then $({\rm{supp}} \hat{f})\bigcup({\rm{supp}} \hat{g})\subseteq D_{+}\bigcup D_{0}$ and $({\rm{supp}} \hat{f})\bigcup({\rm{supp}} \hat{g})\subseteq D_{-}\bigcup D_{0}$ imply ${\rm{supp}} (fg)^{\wedge}\subseteq D_{+}\bigcup D_{0}$ and ${\rm{supp}} (fg)^{\wedge}\subseteq D_{-}\bigcup D_{0}$, respectively.

Proof. The H$\rm\ddot{o}$lder inequality

$\|fg\|_{L^{r}(\mathbb{R}^{n})}\leq\|f\|_{L^{p}(\mathbb{R}^{n})} \|g\|_{L^{q}(\mathbb{R}^{n})}$

(3)

implies that $fg\in L^{r}(\mathbb{R}^{n})$. Thus, for each $\phi\in\mathcal{S}(\mathbb{R}^{n})$ we get that

$\left\langle {{{(fg)}^ \wedge },\phi } \right\rangle = \left\langle {fg,\hat \phi } \right\rangle = \int_{{^n}} f (t)g(t)\hat \phi (t)dt.$

Choose a function $\psi\in\mathcal{D}(\mathbb{R}^{n})$ such that $\hat{\psi}(0)=1$ and set

$f_{j}(x):= \int_{\mathbb{R}^{n}}\psi_{j}(x-t)f(t)dt, \ \ j\in\mathbb{N},$

where $\psi_{j}(t)=j^{n}\psi(jt), t\in\mathbb{R}^{n}.$ For each $j\in\mathbb{N},$ the function $f_{j}$ enjoys the property that $f_{j}\in\mathbb{C}^{\infty}(\mathbb{R}^{n}), D^{k}f_{j}\in L^{\infty}(\mathbb{R}^{n})$ for each nonnegative integer $k$. Furthermore, if $p＜\infty,$ then $f_{j}$ converges to $f$ in $L^{p}(\mathbb{R}^{n})$ as $j$ goes to infinity. This fact as well as $(3)$ yields for $p＜\infty$ that

$\|f_{j}g-fg\|_{L^{r}(\mathbb{R}^{n})}\leq\|f_{j}-f\|_{L^{p}(\mathbb{R}^{n})} \|g\|_{L^{q}(\mathbb{R}^{n})}\rightarrow 0\, \, (j\rightarrow\infty).$

As a result, if $p＜\infty$, then there holds

$\lim_{j\rightarrow\infty}\int_{\mathbb{R}^{n}}f_{j}(t)g(t)\hat{\phi}(t)dt= \int_{\mathbb{R}^{n}}f(t)g(t)\hat{\phi}(t)dt.$

The above equality remains true for $p=\infty$ because in this case we have that $f_{j}\in L^{\infty}(\mathbb{R}^{n})$ converges almost everywhere to $f$ and thus that $g\hat{\phi}\in L^{1}(\mathbb{R}^{n}).$

Now we suppose that $({\rm{supp}} \hat{f})\bigcup({\rm{supp}} \hat{g})\subseteq D_{+}\bigcup D_{0}.$ To show that ${\rm{supp}} (fg)^{\wedge}\subseteq D_{+}\bigcup D_{0}$, it is sufficient to show that $\int_{\mathbb{R}^{n}}f_{j}(t)g(t)\hat{\phi}(t)dt=0$ for each $j\in \mathbb{N}$ and $\phi\in\mathcal{D}_{D_{-}}(\mathbb{R}^{n}).$ For $\phi\in\mathcal{D}_{D_{-}}(\mathbb{R}^{n})$, the properties of the functions $f_{j}$ ensure that $f_{j}\phi\in\mathcal{S}(\mathbb{R}^{n}), j\in \mathbb{N}.$ Therefore,

$\int_{{^n}} g (t){f_j}(t)\hat \phi (t)dt = \left\langle {\hat g,{{({f_j}\hat \phi )}^ \vee }} \right\rangle ,\;j \in \mathbb{N}.$

A direct computation shows that

$\begin{array}{c} {({f_j}\hat \phi )^ \vee }(x) = \smallint_{\mathbb{R}^{n}} \widehat {{\psi _j}}( - t)\hat f( - t)\phi (x - t)dt\\ = \smallint_{\mathbb{R}^{n}} \hat \psi (\frac{{ - t}}{j})\hat f( - t)\phi (x - t)dt\\ = \smallint_{\mathbb{R}^{n}} \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\smile$}} \over \psi } (\frac{{ - t}}{j})\hat f(t)\phi (x + t)dt\\ = \left\langle {\hat f,\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\smile$}} \over \psi } (\frac{ \cdot }{j})\phi (x + \cdot )} \right\rangle . \end{array}$

The above conclusion together with $\phi\in\mathcal{D}_{D_{-}}(\mathbb{R}^{n})$, $({\rm{supp}} \hat{f})\subseteq D_{+}\bigcup D_{0}$ implies that ${\rm{supp}}(f_{j}\hat{\phi})^{\vee}\subseteq D_{-}$. It follows by the assumption $({\rm{supp}} \hat{g})\subseteq D_{+}\bigcup D_{0}$ that

$\left\langle {\hat g,{{({f_j}\hat \phi )}^ \vee }} \right\rangle = 0,\;\;j \in \mathbb{N}.$

Therefore, we get ${\rm{supp}} (fg)^{\hat{}}\subseteq D_{+}\bigcup D_{0}$. The other case can be dealt with likewise. This immediately completes the proof.

Lemma 2.4 If $f\in L^{p}(\mathbb{R}^{n})$ for $1＜p＜\infty,$ then ${\rm{supp}}(f-(-i)^{n}Hf)^{\wedge}\subseteq D_{-}\bigcup D_{0}.$

Proof. According to the definition, we need to show for each $\phi\in\mathcal{D}_{D_{+}}(\mathbb{R}^{n})$ that

$\left\langle {{{(f - {{( - i)}^n}Hf)}^ \wedge },\phi } \right\rangle = 0.$

For this purpose, we point out that if $1\leq p＜\infty$ then for each $\psi\in\mathcal{S}(\mathbb{R}^{n})$ there holds

$\left\langle {Hf,\psi } \right\rangle = \left\langle {f,{{( - 1)}^n}H\psi } \right\rangle .$

(4)

Let $\phi\in\mathcal{D}_{D_{+}}(\mathbb{R}^{n})$, then

$\begin{array}{c} \left\langle {{{(f - {{( - i)}^n}Hf)}^ \wedge },\phi } \right\rangle = \left\langle {(f - {{( - i)}^n}Hf),\hat \phi } \right\rangle \\ = \left\langle {f,\hat \phi - {{(i)}^n}H\hat \phi } \right\rangle \\ = \left\langle {\hat f,\phi - {{( - 1)}^n}sgn( \cdot )\phi } \right\rangle \\ = 0, \end{array}$

thereby completing the proof.

Lemma 2.5 Let $f\in L^{p}(\mathbb{R}^{n})$ for $1＜p＜\infty$, then on $D_{+}\bigcup D_{-}$, ${\rm{supp}}\hat{f}\subseteq D_{+}\bigcup D_{0}$ and ${\rm{supp}}\hat{f}\subseteq D_{-}\bigcup D_{0}$ are equivalent to $Hf=(i)^{n}f$ and $Hf=-(i)^{n}f$, respectively.

Proof. Suppose $f\in L^{p}(\mathbb{R}^{n})$ satisfies that $Hf=(i)^{n}f$. Then we have for each $\phi\in\mathcal{D}_{D_{-}}(\mathbb{R}^{n})$ that

$\left\langle {\hat f,\phi } \right\rangle = \left\langle {f,\hat \phi } \right\rangle = \frac{1}{2}(\left\langle {f,\hat \phi } \right\rangle + {( - i)^n}\left\langle {{{(i)}^n}f,\hat \phi } \right\rangle ) = \frac{1}{2}(\left\langle {f,\hat \phi } \right\rangle + {( - i)^n}\left\langle {Hf,\hat \phi } \right\rangle ).$

(5)

According to $(4)$, we get that

$\left\langle {f,\hat \phi } \right\rangle + {( - i)^n}\left\langle {Hf,\hat \phi } \right\rangle = \left\langle {f,\hat \phi + {{(i)}^n}H\hat \phi } \right\rangle = \left\langle {\hat f,\phi (1 + {{( - 1)}^n}sgn( \cdot ))} \right\rangle = 0.$

(6)

Combining $(5)$ with $(6)$ proves that $\left\langle {\hat f,\phi } \right\rangle = 0$ for each $\phi\in\mathcal{D}_{D_{-}}(\mathbb{R}^{n})$, that is ${\rm{supp}}\hat{f}\subseteq D_{+}\bigcup D_{0}$.

Conversely, suppose that we have $f\in L^{p}(\mathbb{R}^{n})$ with ${\rm{supp}}\hat{f}\subseteq D_{+}\bigcup D_{0}$. By lemma $\ref{lem4}$, there holds ${\rm{supp}}(f+(-i)^{n}Hf)^{\wedge}\subseteq D_{+}\bigcup D_{0}$, then ${\rm{supp}}{(Hf)}^{\wedge}\subseteq D_{+}\bigcup D_{0}$. Therefore, to show that $Hf=(i)^{n}f$ it suffices to show for each $\phi\in\mathcal{D}_{D_{+}}(\mathbb{R}^{n})$ that

$\left\langle {{{(f - {{( - i)}^n}Hf)}^ \wedge },\phi } \right\rangle = 0.$

Lemma 2.4 work for this purpose. This ends the proof.

Now we introduce that each $f\in L^{p}(\mathbb{R}^{n})$ for $1＜p＜\infty$ can be decomposed as $f=f_{+}+f_{-}$, where

$f_{+}=\frac{f+(-i)^{n}Hf}{2}, \quad f_{-}=\frac{f-(-i)^{n}Hf}{2}.$

Theorem 2.6 Let $f\in L^{p}(\mathbb{R}^{n}), g\in L^{q}(\mathbb{R}^{n})$ with $\frac{1}{p}+\frac{1}{q}=\frac{1}{r}.$ Then $f, g$ satisfy the Bedrosian identity $H(fg)=fHg$ on $D_{+}\bigcup D_{-}$ if and only if

${\rm{supp}}(f_{+}g_{-})^{\wedge}\subseteq D_{-}\bigcup D_{0}\ {\rm{and}}\ \ {\rm{supp}}(f_{-}g_{+})^{\wedge}\subseteq D_{+}\bigcup D_{0}.$

@Proof. By using the above decomposition $f=f_{+}+f_{-}$ and $g=g_{+}+g_{-},$ we can rewrite the Bedrosian identity as

$H(f_{+}g_{+}+f_{-}g_{-}+f_{+}g_{-}+f_{-}g_{+})=f_{+}Hg_{+}+f_{-}Hg_{-} +f_{+}Hg_{-}+f_{-}Hg_{+}.$

(7)

For each $\phi\in L^{q}(\mathbb{R}^{n})$, it holds

$\begin{array}{c} \left\langle {H{f_ + },\phi } \right\rangle = \left\langle {{f_ + },{{( - 1)}^n}H\phi } \right\rangle \\ = \left\langle {\frac{{f + {{( - i)}^n}Hf}}{2},{{( - 1)}^n}H\phi } \right\rangle \\ = \frac{1}{2}\left\langle {f,{{( - 1)}^n}H\phi } \right\rangle + \frac{{{{( - i)}^n}}}{2}\left\langle {f,{H^2}\phi } \right\rangle \\ = \frac{1}{2}\left\langle {Hf,\phi } \right\rangle + \frac{{{{( - i)}^n}}}{2}\left\langle {f,{H^2}\phi } \right\rangle \\ = \left\langle {{{(i)}^n}\frac{{f + {{( - i)}^n}Hf}}{2},\phi } \right\rangle \\ = \left\langle {{{(i)}^n}{f_ + },\phi } \right\rangle . \end{array}$

That is $Hf_{+}=(i)^{n}f_{+}.$ Adopting the same argument, one may conclude that $Hf_{-}=-(i)^{n}f_{-}.$ The above fact leads us to the following equivalent of $(6)$

$H(f_{+}g_{+}+f_{-}g_{-}+f_{+}g_{-}+f_{-}g_{+})= (i)^{n}f_{+}g_{+}-(i)^{n}f_{-}g_{-}+(i)^{n}f_{-}g_{+}-(i)^{n}f_{+}g_{-}.$

(8)

According to Lemma 2.5, it is easy to check that

${\rm{supp}}\hat{f}_{+}\subseteq D_{+}\bigcup D_{0}, \ \ {\rm{supp}}\hat{f}_{-}\subseteq D_{-}\bigcup D_{0},$

and similarly

${\rm{supp}}\hat{g}_{+}\subseteq D_{+}\bigcup D_{0}, \ \ {\rm{supp}}\hat{g}_{-}\subseteq D_{-}\bigcup D_{0}.$

An application of Lemma 2.3 then yields that

${\rm{supp}}(f_{+}g_{+})^{\wedge}\subseteq D_{+}\bigcup D_{0}, \ \ {\rm{supp}}(f_{-}g_{-})^{\wedge}\subseteq D_{-}\bigcup D_{0}.$

(9)

By the H$\rm\ddot{o}$lder inequality, functions $f_{+}g_{+}, f_{+}g_{-}, f_{-}g_{+}, f_{-}g_{-}$ are all in $L^{r}(\mathbb{R}^{n})$. The results relation (8) hence imply by Lemma 2.5 that

$H(f_{+}g_{+})=(i)^{n}f_{+}g_{+}, \ \ H(f_{-}g_{-})=-(i)^{n}f_{-}g_{-}.$

Therefore, equation (7) holds if and only if

$H(f_{+}g_{-}+f_{-}g_{+})=(i)^{n}f_{-}g_{+}-(i)^{n}f_{+}g_{-}.$

(10)

If ${\rm{supp}}(f_{+}g_{-})^{\wedge}\subseteq D_{-}\bigcup D_{0}\ {\rm{and}}\ \ {\rm{supp}}(f_{-}g_{+})^{\wedge}\subseteq D_{+}\bigcup D_{0}$ holds true, then (9) is valid. On the other hand, we suppose that (9) is true. By applying the Hilbert transform to both sides of equation (9), one gets that

$(-1)^{n}(f_{+}g_{-}+f_{-}g_{+})=(i)^{n}H(f_{-}g_{+})-(i)H(f_{+}g_{-}).$

(11)

Combining $(10)$ and $(9)$, it follows that

$H(f_{-}g_{+})=(i)^{n}f_{-}g_{+}, \ \ H(f_{+}g_{-})=-(i)^{n}f_{+}g_{-}.$

The above is equivalent to ${\rm{supp}}(f_{+}g_{-})^{\wedge}\subseteq D_{-}\bigcup D_{0}\ {\rm{and}} {\rm{supp}}(f_{-}g_{+})^{\wedge}\subseteq D_{+}\bigcup D_{0}$. The proof is complete.

Theorem 2.7. If $f\in L^{p}(\mathbb{R}^{n}), g\in L^{q}(\mathbb{R}^{n})$ satisfy either $({\rm{supp}} \hat{f})\bigcup({\rm{supp}}\hat{g})\subseteq D_{+}\bigcup D_{0}$ or $({\rm{supp}} \hat{f})\bigcup({\rm{supp}}\hat{g})\subseteq D_{-}\bigcup D_{0}$ then the Bedrosian identity holds on $D_{+}\bigcup D_{-}$ .

Proof. If $({\rm{supp}} \hat{f})\bigcup({\rm{supp}}\hat{g})\subseteq D_{+}\bigcup D_{0}$, then by Lemma 2.5, $Hf=(i)^{n}f$ and $Hg=-(i)^{n}g$. Therefore, according to the definition of $f_{+}$ and $f_{-}$, we have that

$f_{+}=\frac{f+(-i)^{n}Hf}{2}=\frac{f+(-i)^{n}(i)^{n}f}{2}=f, \quad f_{-}=0.$

Similarly $g_{+}=g, \ g_{-}=0$. The desired result then follows by a trivial application of Theorem 2.6. The other case can be proved in the same way. The proof is thus completed.

Theorem 2.8. Suppose $f\in \mathcal{S}(\mathbb{R}^{n}), g\in L^{2}(\mathbb{R}^{n})$ with $\frac{1}{p}+\frac{1}{q}=1$ satisfy the Bedrosian identity if and only if

$2\int_{D_{-}}\hat{f}(\xi-\eta)\hat{g}(\eta)d\eta+ \int_{D_{0}}\hat{f}(\xi-\eta)\hat{g}(\eta)d\eta=0, \ \ \xi\in D_{+}\bigcup D_{0}$

(12)

and

$2\int_{D_{+}}\hat{f}(\xi-\eta)\hat{g}(\eta)d\eta+ \int_{D_{0}}\hat{f}(\xi-\eta)\hat{g}(\eta)d\eta=0, \ \ \xi\in D_{-}\bigcup D_{0}.$

(13)

Proof. Since $fg, fHg\in L^{1}(\mathbb{R}^{n}),$ $H(fg)=fHg\ a.e.$ if and only if

$(fHg)^{\wedge}(\xi)=(-i)^{n}sgn(\xi)(fg)^{\wedge}(\xi), \ \xi\in \mathbb{R}^{n}.$

(14)

The equation (13) admits the form

$(-i)^{n}sgn(\xi)\int_{\mathbb{R}^{n}}\hat{f}(\xi-\eta)\hat{g}(\eta)d\eta= \int_{\mathbb{R}^{n}}\hat{f}(\xi-\eta)(-i)^{n}sgn(\eta)\hat{g}(\eta)d\eta,$

which gives

$\int_{\mathbb{R}^{n}}\hat{f}(\xi-\eta)\hat{g}(\eta)(sgn(\xi)-sgn(\eta))d\eta=0, \ \xi\in \mathbb{R}^{n}.$

Clearly, the above integral can be divided into the following three parts

$2\int_{D_{-}}\hat{f}(\xi-\eta)\hat{g}(\eta)d\eta+ \int_{D_{0}}\hat{f}(\xi-\eta)\hat{g}(\eta)d\eta=0, \ \ \xi\in D_{+},$

$2\int_{D_{+}}\hat{f}(\xi-\eta)\hat{g}(\eta)d\eta+ \int_{D_{0}}\hat{f}(\xi-\eta)\hat{g}(\eta)d\eta=0, \ \ \xi\in D_{-},$

$\int_{D_{-}}\hat{f}(\xi-\eta)\hat{g}(\eta)d\eta= \int_{D_{+}}\hat{f}(\xi-\eta)\hat{g}(\eta)d\eta, \ \ \ \xi\in D_{0}.$

By the continuity of $\int_{D_{+}}\hat{f}(\cdot-\eta)\hat{g}(\eta)d\eta$ and $\int_{D_{-}}\hat{f}(\cdot-\eta)\hat{g}(\eta)d\eta$, it is not hard to show that $f$ and $g$ satisfy the three integrals above if and only they satisfy (11) and (12). We thus conclude the proof.

3. Distribution $f\in L^{p}(\mathbb{R}^{n})$ and the Hardy space on tube

In this section, we give some lemmas which will be used in the proof process of the main result.

Lemma 3.1. (see [10]) Let $u\in L^{p}(\mathbb{R})$ for $1<p<\infty$. The function $G(u)(z)$ is defined as

$G(u)(z)=\frac{1}{\pi i}\int_{\mathbb{R}}\frac{u(t)}{t-z}dt.$

Then $G(u)\in H^{p}(\mathbb{C}_{+}).$ Moreover,

$\int_{\mathbb{R}}|G(u)(x+iy)|^{p}dx\leq A_{p}\int_{\mathbb{R}}|u(t)|^{p}dt,$

where $A_{p}=\max\{\frac{p2^{p}}{p-1}, 2^{p}p^{p-1}\}.$

Lemma 3.2. Let $\Gamma=\{y=(y_{1}, y_{2})\in\mathbb{R}^{2}:y_{1}>0, y_{2}>0\}$. Suppose $f\in L^{p}(\mathbb{R}^{2})\, \, (1<p<\infty)$. The Cauchy integral of $f$ is given by

$F(z)=\frac{1}{(2\pi i)^{2}}\int_{\mathbb{R}^{2}}K(z-t)f(t)dt.$

Then $F(z)\in H^{p}(T_{\Gamma})$. Moreover, there exists a constant $A$ such that

$\|F\|_{H^{p}(T_{\Gamma})}\leq A\|f\|_{p}.$

Proof. We consider the first octant $\Gamma=\{y=(y_{1}, y_{2})\in\mathbb{R}^{n}:y_{1}>0, y_{2}>0\}$, then we can show $\Gamma^{\ast}=\overline{\Gamma}$ ($\overline{\Gamma}$ is the closure of $\Gamma$). The Cauchy kernel associated with the tube $T_{\Gamma}=\{z=x+iy:x\in\mathbb{R}^{2}, y\in\Gamma\}$ is

$K(z)=\frac{1}{(2\pi i)^{2}}\int_{\Gamma^{\ast}}e^{2\pi iz\cdot t}dt=\int_{0}^{\infty}\int_{0}^{\infty} e^{2\pi i(z_{1}t_{1}+z_{2}t_{2})}dt_{1} dt_{2}=\prod_{j=1}^{2}\frac{-1}{2\pi iz_{j}}.$

Direct calculation yields that

$F(z)=\frac{1}{(2\pi i)^{2}}\int_{\mathbb{R}^{2}}K(z-t)f(t)dt\nonumber\\ =\frac{1}{(2\pi i)^{2}}\int_{\mathbb{R}}\int_{\mathbb{R}}\frac{f(t_{1}, t_{2})} {(t_{1}-z_{1})(t_{2}-z_{2})}dt_{1} d t_{2}.$

Now we denote

$(Cf)(z_{1}, t_{2})=\int_{\mathbb{R}}\frac{f(t_{1}, t_{2})}{t_{1}-z_{1}}dt_{1}.$

By using Lemma 3.1. as well as the fact $f\in L^{p}(\mathbb{R}^{2})$, we know that

$(Cf)_{t_{2}}(z_{1})=(Cf)(z_{1}, t_{2})\in H^{p} (\mathbb{C}_{+}),$

and

$\int_{\mathbb{R}}|(Cf)_{t_{2}}(x_{1}+iy_{1})|^{p}dx_{1}\leq A_{p}\int_{\mathbb{R}}|f(t_{1}, t_{2})|^{p}dt_{1}<\infty,$

where $A_{p}$ is a constant and $\mathbb{C}_{+}=\{z=x+iy:x\in\mathbb{R}, y>0\}$ is the upper half-plane in $\mathbb{C}$. It thus gives

$\int_{\mathbb{R}}\int_{\mathbb{R}}|(Cf)(z_{1}, t_{2})|^{p}dx_{1}dt_{2}\leq A_{p}\int_{\mathbb{R}}\int_{\mathbb{R}}|f(t_{1}, t_{2})|^{p}dt_{1}dt_{2}<\infty.$

As a result, we have $(Cf)_{z_{1}}(t_{2})=(Cf)(z_{1}, t_{2})\in L^{p}(\mathbb{R}).$ Keeping in mind the following fact

$F(z)=\frac{1}{(2\pi i)^{2}}\int_{\mathbb{R}}\int_{\mathbb{R}}\frac{f(t_{1}, t_{2})} {(t_{1}-z_{1})(t_{2}-z_{2})}dt_{1} d t_{2}=\int_{\mathbb{R}}\frac{(Cf)_{z_{1}}(t_{2})}{t_{2}-z_{2}}dt_{2},$

as well as lemma 2.1, we can show that for such fixed $z_{1}$

$\int_{\mathbb{R}}\frac{(Cf)_{z_{1}}(t_{2})}{t_{2}-z_{2}}dt_{2}\in H^{p}(\mathbb{C}_{+}).$

Thus, it gives

$\int_{\mathbb{R}}\Big|\int_{\mathbb{R}}\frac{(Cf)_{z_{1}}(t_{2})}{t_{2}-z_{2}} dt_{2}\Big|^{p}dx_{2}\leq A_{p}\int_{\mathbb{R}}|(Cf)_{z_{1}}(t_{2})|^{p}dt_{2}.$

Then, we can conclude

$\int_{\mathbb{R}}\int_{\mathbb{R}}\Big|\int_{\mathbb{R}}\frac{(Cf)_{z_{1}}(t_{2})}{t_{2}-z_{2}} dt_{2}\Big|^{p}dx_{2}dx_{1} \leq A_{p}\int_{\mathbb{R}}\int_{\mathbb{R}}|(Cf)_{z_{1}}(t_{2})|^{p}dt_{2}dx_{1}\nonumber\\ \leq A_{p}^{2}\int_{\mathbb{R}}\int_{\mathbb{R}}|f(t_{1}, t_{2})|^{p}dt_{1}dt_{2}\nonumber\\ < \infty.$

Therefore, $F(z)\in H^{p}(T_{\Gamma})$. Moreover, there exists a constant $A$ such that

$\|F\|_{H^p}\leq A\|f\|_{p}.$

Thus, we complete the proof.

Remark 3.3. Adopting the induction, we can get the same conclusion when n is a finite and positive integer.

Corollary 3.4. Let $\Gamma$ be an open cone in $\mathbb{R}^{n}$ and $F\in H^{p}(T_{\Gamma})$ for $1\leq p\leq2$, then $F(z)$ has the form

$F(z)=\int_{\mathbb{R}^{n}}K(z-\xi)F(\xi)d\xi,$

where $F(\xi)=\lim_{\eta → 0 \eta\in\Gamma}F(\xi+i\eta)\ $ in $L^{p}(\mathbb{R}^{n})$.

Proof. By using [12, Theorem 3.3.5]{Fei}, the above Corollary follows.

With these lemmas at our dispersal, we move to prove the main results.

Theorem 3.5. Let $f\in\mathcal{D}_{L^{p}}'(\mathbb{R}^{n})$ for $1<p<\infty$. We set

$F(z)=\frac{1}{(2\pi i)^{n}} \Big<f(t), \, \, \frac{1}{\prod_{i=1}^{n}(t_{i}-z_{i})}\Big>,$

then

$|F(x+iy)|^{p}\leq\frac{A_{\delta}}{|y_{1}y_{2}\cdots y_{n}|}, \quad |y_{i}|>\delta_{i}\ (i=1, 2, \cdots, n),$

where $A_{\delta}$ is a constant, $\delta=(\delta_{1}, \cdots, \delta_{n}), \ \delta_{i}>0\ (i=1, 2\cdots, n).$ Moreover, there exist nonnegative integer $j$, and functions $F_{\alpha}(z)\in H^{p}(T_{\Gamma})$ such that

$F(z)=\sum_{|\alpha|\leq j}D^{\alpha}F_{\alpha}(z),$

where $\alpha=(\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n})$ is a multi-index notation and $\Gamma=\{y=(y_{1}, y_{2}, \cdots, y_{n}\in \mathbb{R}^{n}):y_{1}>0, y_{2}>0, \cdots, y_{n}>0\}$ is the first octant.

Proof. For $f\in \mathcal{D}_{L^{p}}^{\prime}(\mathbb{R}^{n})$, $1<p<\infty$, by Structure Formula [27] there exists a function $g_{\alpha}\in L^{p}(\mathbb{R}^{n})$, such that the distribution $f$ admits the form

$f=\Sigma_{|\alpha|\leq j}D^{\alpha}g_{\alpha}.$

Therefore, we have

$\begin{array}{l} |F(z)| = |\frac{1}{{{{(2\pi i)}^n}}}\left\langle {\sum\limits_{|\alpha | \le j} {D_t^\alpha } {g_\alpha },\frac{1}{{\prod\limits_{i = 1}^n {({t_i} - {z_i})} }}} \right\rangle |\\ \le \frac{1}{{{{(2\pi )}^n}}}\sum\limits_{|\alpha | \le j} {\int_{{^n}} | } {g_\alpha }|{\mkern 1mu} {\mkern 1mu} |\prod\limits_{i = 1}^n {\frac{{{\alpha _i}!}}{{{{({t_i} - {z_i})}^{{\alpha _i} + 1}}}}} |dt\\ \le \frac{1}{{{{(2\pi )}^n}}}\sum\limits_{|\alpha | \le j} [ {g_\alpha }{_p}\prod\limits_{i = 1}^n {({\alpha _i}!)} {(\int_\mathbb{R} {\frac{1}{{{{(t_i^2 + 1)}^{\frac{{p'({\alpha _i} + 1)}}{2}}}|{y_i}{|^{({\alpha _i} + 1)p' - 1}}}}} d{t_i})^{\frac{1}{{p'}}}}]\\ = \frac{1}{{|{y_1}{y_2} \cdots {y_n}{|^{\frac{1}{p}}}}}\frac{1}{{{{(2\pi )}^n}}}\sum\limits_{|\alpha | \le j} ( {g_\alpha }{_p}\prod\limits_{i = 1}^n {({\alpha _i}!)} {(\int_ {\frac{1}{{{{(t_i^2 + 1)}^{\frac{{p'({\alpha _i} + 1)}}{2}}}}}} \frac{1}{{|{y_i}{|^{{\alpha _i}}}}}d{t_i})^{\frac{1}{{p'}}}})\\ \le \frac{{{B_\delta }}}{{|{y_1}{y_2} \cdots {y_n}{|^{\frac{1}{p}}}}}, \end{array}$

where

$B_{\delta}=\frac{1}{|y_{1}y_{2}\cdots y_{n}| ^{\frac{1}{p}}} \frac {1} {(2\pi )^{n}}\sum_{|\alpha|\leq j}\left(\|g_{\alpha}\|_{p}\prod_{i=1}^{n}(\alpha_{i}!)\Big(\int_{\mathbb{R}} \frac{1}{(t_{i}^{2}+1)^{\frac {p'(\alpha_{i}+1)}{2}}}\frac{1}{|\delta_{i}|^{\alpha_{i}}}dt_{i} \Big)^{\frac{1}{p'}}\right),$

$|y_{i}|\geq \delta_{i}$, $\alpha_{i}\geq0,$ $(i=1, 2, \cdots, n), \ |\alpha|=\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}, \ D_{x}^{\alpha}=\frac{\partial^{|\alpha|}} {\partial x_{1}^{\alpha_{1}}\cdots\partial x_{n}^{\alpha_{n}}}$ and $\frac{1}{p}+\frac{1}{p'}=1.$ Denoting $A_{\delta}=B_{\delta}^{p}$, we get

$|F(x+iy)|^{p}\leq\frac{A_{\delta}}{|y_{1}y_{2}\cdots y_{n}|}, \ \ \ \ |y_{i}|\geq \delta_{i}, \ \ \ i=1, 2, \cdots, n.$

Now we start to compute $F(z)$:

$\begin{eqnarray} F(z)=\frac{1}{(2\pi i)^{n}}\Big<\sum_{|\alpha|\leq j}D_{t}^{\alpha}g_{\alpha}, \, \, \frac{1}{\prod_{i=1}^{n}(t_{i}-z_{i})}\Big>\nonumber\\ =\frac{1}{(2\pi i)^{n}}\sum_{|\alpha|\leq j}\int_{\mathbb{R}^{n}}g_{\alpha}(t) D_{z}^{\alpha}\Big(\frac{1}{\prod_{i=1}^{n}(t_{i}-z_{i})}\Big)dt\nonumber\\ =\sum_{|\alpha|\leq j}D_{z}^{\alpha} F_{\alpha}(z), \nonumber\ \end{eqnarray}$

where

$F_{\alpha}(z)=\frac{1}{(2\pi i)^{n}} \int_{\mathbb{R}^{n}}\frac{g_{\alpha}(t)}{\prod_{i=1}^{n}(t_{i}-z_{i})}dt.$

In view of lemma 3.1 and Remark 3.3, we have

$F_{\alpha}(z)\in H^{p}(T_{\Gamma}).$

This completes the proof of Theorem 3.5.

Corollary 3.6. Let $1<p\leq2$. Assume that $j$ is a nonnegative integer and $F_{\alpha}(z)\in H^p(T_{\Gamma})$. We denote

$F(z)=\sum_{|\alpha|\leq j}D^{\alpha}F_{\alpha}(z),$

then there exists $f(x)\in \mathcal{D}'_{L^{p}}(\mathbb{R}^{n})$ such that

$F(z)=\frac{1}{(2\pi i)^{n}} \Big<f(t), \, \, \frac{1}{\prod_{i=1}^{n}(t_{i}-z_{i})}\Big>,$

where $\alpha$ is a multi-index notation and $\Gamma$ is an open convex cone.

Acknowledgments

The authors would like to thank the anonymous referee and editor very much for their valuable comments and suggestions, which greatly help us improve the presentation of this article. This work was supported by SRFDP (Grant 20100003110004) and by NSFC (Grant 11271045).

Conflict of Interest

We declare that we have no conflict of interest.