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# New Poisson inequality for the Radon transform of infinitely differentiable functions

Journal of Inequalities and Applications20182018:214

https://doi.org/10.1186/s13660-018-1805-9

• Accepted: 6 August 2018
• Published:

## Abstract

Poisson inequality for the Radon transform is a key tool in signal analysis and processing. An analogue of the Hardy–Littlewood–Poisson inequality for the Radon transform of infinitely differentiable functions is proved. The result is related to a paper of Luan and Vieira (J. Inequal. Appl. 2017:12, 2017) and to a previous paper by Yang and Ren (Proc. Indian Acad. Sci. Math. Sci. 124(2):175-178, 2014).

## Keywords

• Poisson inequality
• Infinitely differentiable functions

## 1 Introduction

The Radon transform $$\mathfrak{PI}$$, which is defined as the Cauchy principal value of the following singular integral
$$(\mathfrak{PI}h) (x):=p.v.\frac{1}{\pi } \int_{\mathbb{R}} \frac{h(y)}{x-y}\,dy=\lim_{\epsilon \to 0} \int_{\vert y-x\vert \ge \epsilon } \frac{h(y)}{x-y}\,dy$$
for any $$x\in \mathbb{R}$$, has been widely used in physics, engineering, and mathematics. The following Poisson inequality
$$\mathfrak{PI}(hg)\leq h\mathfrak{PI}g$$
(1.1)
was first studied in [13, 5]. It was proved that (1.1) holds if $$h,g \in L^{2}(\mathbb{R})$$ satisfy that $$\operatorname{supp}\hat{f}\subseteq \mathbb{R}_{+}$$ ($$\mathbb{R}_{+}=[0,\infty)$$) and $$\operatorname{supp}\hat{g}\subseteq \mathbb{R}_{+}$$ in [21].

In 2014, Yang and Ren also obtained more general sufficient conditions by weakening the above condition in [24]. Recently, Luan and Vieria established the first necessary and sufficient condition in the time domain and a parallel result in the frequency domain for the Poisson inequality in [16].

It is natural that there have been attempts to define the complex signal and prove the Poisson inequality in a multidimensional case.

### Definition 1.1

The partial Radon transform $$\mathfrak{PI}_{j}$$ of a function $$h \in L^{p}(\mathbb{R}^{n})$$ ($$1 \leq p < \infty$$) is given by
$$(\mathfrak{PI}_{j}h) (x):=p.v.\frac{1}{\pi } \int_{\mathbb{R}}\frac{h(y)}{x _{j}-y_{j}}\,dy_{j}.$$
The total Radon transform $$\mathfrak{PI}$$ of a function $$h \in L^{p}( \mathbb{R}^{n})$$ ($$1 \le p< \infty$$) is defined as follows:
\begin{aligned} (\mathfrak{PI}h) (x) :=&p.v.\frac{1}{\pi^{n}} \int_{\mathbb{R}^{n}}\frac{h(y)}{ \Pi^{n}_{j=1}(x_{j}-y_{j})}\,dy \\ =&\lim_{max \epsilon_{j}\to 0} \int_{\vert y_{j}-x_{j}\vert \ge \epsilon_{j}>0,j=1,2,\ldots, n}\frac{h(y)}{\Pi^{n} _{j=1}(x_{j}-y_{j})}\,dy. \end{aligned}
The existence of the singular integral above and its boundedness property
$$\Vert \mathfrak{PI}h\Vert _{L^{p}(\mathbb{R}^{n})}\le C^{n}_{p} \Vert h\Vert _{L^{p}( \mathbb{R}^{n})}$$
were proved in [10, 19]. The iterative nature of the Radon transform in $$L^{p}(\mathbb{R}^{n})$$ ($$p > 1$$) was shown in [6]. It was shown that
$$\mathfrak{PI}=\prod^{n}_{j=1} \mathfrak{PI}_{j}.$$
The operations $$\mathfrak{PI}_{i}$$ and $$\mathfrak{PI}_{j}$$ commute with each other, where $$i,j = 1,2,\ldots, n$$.
It is known that the Fourier transform ĥ of $$h \in L^{1}( \mathbb{R}^{n})$$ is defined as follows (see [7]):
$$\hat{h}(x)= \int_{\mathbb{R}^{n}}h(t)e^{-ix.t}\,dt,$$
where $$x\in \mathbb{R}^{n}$$.

Let $$\mathcal{D}(\mathbb{R}^{n})$$ be the space of infinitely differentiable functions in $$\mathbb{R}^{n}$$ with a compact support and $$\mathcal{D}^{\prime }(\mathbb{R}^{n})$$ be the space of distributions, that is, the dual of $$\mathcal{D}(\mathbb{R}^{n})$$ (see [15, 23]). This definition is consistent with the ordinary one when T is a continuous function.

Put
\begin{aligned}& D_{+}=\Biggl\{ x:x\in \mathbb{R}^{n}, \operatorname{sgn}(-x)=\prod ^{n}_{j=1}\operatorname{sgn}(-x_{j})=1 \Biggr\} , \\& D_{-}=\Biggl\{ x:x\in \mathbb{R}^{n}, \operatorname{sgn}(-x)=\prod ^{n}_{j=1}\operatorname{sgn}(-x_{j})=-1 \Biggr\} , \end{aligned}
and
$$D_{0}=\Biggl\{ x:x\in \mathbb{R}^{n}, \operatorname{sgn}(-x)=\prod ^{n}_{j=1}\operatorname{sgn}(-x_{j})=0 \Biggr\} .$$

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.

The Schwartz class $$\mathcal{S}(\mathbb{R}^{n})$$ consists of all infinitely differentiable functions φ on $$\mathbb{R}^{n}$$ satisfying
$$\sup_{x\in \mathbb{R}^{n}}\bigl\vert x^{\alpha }D^{\beta }\varphi (x)\bigr\vert < \infty$$
for all $$\alpha,\beta \in \mathbb{Z}^{n}_{+}$$, where $$\alpha =(\alpha _{1},\alpha_{2},\ldots,\alpha_{n})$$, $$\beta =(\beta_{1},\beta_{2}, \ldots,\beta_{n})$$, $$\alpha_{j}$$ ($$j=1,2,\ldots,n$$) and $$\beta_{j}$$ ($$j=1,2, \ldots,n$$) are nonnegative integers.
The Fourier transform φ̂ is a linear homeomorphism from $$S(\mathbb{R}^{n})$$ onto itself. Meanwhile, the following identity holds:
$$(\mathfrak{PI}\varphi)^{\wedge }(x)=(-i)\operatorname{sgn}(x)\hat{\varphi },$$
where $$\varphi \in \mathcal{S}(\mathbb{R}^{n})$$.
The Fourier transform $$F:\mathbb{S}^{\prime }(\mathbb{R}^{n}) \to \mathbb{S}^{\prime }(\mathbb{R}^{n})$$ defined as
$$\langle \hat{\nu },\varphi \rangle = \langle \nu, \hat{\varphi } \rangle$$
for any $$\varphi \in \mathcal{S}(\mathbb{R}^{n})$$ is a linear isomorphism from $$\mathbb{S}^{\prime }(\mathbb{R}^{n})$$ onto itself. For the detailed properties of $$\mathbb{S}(\mathbb{R}^{n})$$ and $$\mathbb{S}^{\prime }(\mathbb{R}^{n})$$, we refer the readers to [18, 20].
For $$\nu \in \mathcal{S}^{\prime }(\mathbb{R}^{n})$$ and $$\varphi \in \mathcal{S}(\mathbb{R}^{n})$$, it is obvious that
$$\langle \tilde{\breve{\nu }},\varphi \rangle = \langle \tilde{\nu }, \hat{ \varphi } \rangle = \langle \nu, \breve{\tilde{\varphi }} \rangle = \langle \hat{ \nu }, \varphi \rangle = \langle \nu,\hat{\varphi } \rangle$$
for any $$\varphi \in \mathcal{S}(\mathbb{R}^{n})$$, where
$$\tilde{\varphi }(x)=\varphi (-x)$$
and ν̃ is defined as follows:
$$\langle \breve{\nu },\varphi \rangle = \langle \nu, \tilde{\varphi } \rangle.$$
So we obtain that
$$\tilde{\breve{\nu }}=\hat{\nu }$$
in the distributional sense.
Following the definition in [16], a function φ belongs to the space $$\mathcal{D}_{L^{p}}(\mathbb{R}^{n})$$ ($$1\le p<\infty$$) if and only if
1. (I)

$$\varphi \in C^{\infty }(\mathbb{R}^{n})$$;

2. (II)
$$D^{k}\varphi \in L^{p}(\mathbb{R}^{n})$$ ($$k=1,2,\ldots$$), where $$C^{\infty }(\mathbb{R}^{n})$$ consists of infinitely differentiable functions,
$$D^{k}\varphi (x)=\frac{\partial^{\vert k\vert }}{\partial x^{k_{1}}_{1}\cdots \partial x^{k_{n}}_{n}}\varphi (x),$$
where $$|k|=k_{1}+k_{2}+\cdots +k_{n}$$ and $$k=(k_{1},k_{2},\ldots,k _{n})$$.

In the sequel, we denote by $$\mathcal{D}^{\prime }_{L^{p}}(\mathbb{R} ^{n})$$ the dual of the corresponding spaces
$$\mathcal{D}_{L^{p^{\prime }}}\bigl(\mathbb{R}^{n}\bigr),$$
where
$$\frac{1}{p}+\frac{1}{p^{\prime }}=1.$$
As a consequence, we have
$$\mathcal{D}\bigl(\mathbb{R}^{n}\bigr)\subseteq \mathcal{S}\bigl( \mathbb{R}^{n}\bigr) \subseteq \mathcal{D}_{L^{p}}\bigl( \mathbb{R}^{n}\bigr)\subseteq L^{p}\bigl( \mathbb{R}^{n}\bigr)$$
and
$$L^{p}\bigl(\mathbb{R}^{n}\bigr)\subseteq \mathcal{D}^{\prime }_{L^{p}}\bigl( \mathbb{R}^{n}\bigr) \subseteq \mathcal{S}^{\prime }\bigl(\mathbb{R}^{n}\bigr)\subseteq \mathcal{D}^{\prime }\bigl(\mathbb{R}^{n}\bigr).$$

### Definition 1.2

Let $$h\in \mathcal{D}^{\prime }_{L^{p}}(\mathbb{R}^{n})$$, where $$1< p<\infty$$. Then the Radon transform of h is defined by (see [8])
$$\langle \mathfrak{PI}h,\varphi \rangle = \bigl\langle f,(-1)^{n} \mathfrak{PI}\varphi \bigr\rangle$$
for any $$\varphi \in \mathcal{D}_{L^{p^{\prime }}}(\mathbb{R}^{n})$$.

In [16], Luan and Vieira proved that the total Radon transform is a linear homeomorphism from $$\mathcal{D}_{L^{p}}(\mathbb{R}^{n})$$ onto itself, as well as if $$h\in \mathcal{D}^{\prime }_{L^{p}}( \mathbb{R}^{n})$$ ($$1< p<\infty$$), then $$\mathfrak{PI}h\in \mathcal{D} ^{\prime }_{L^{p}}(\mathbb{R}^{n})$$ and the Radon transform H defined above is a linear isomorphism from $$\mathcal{D}^{\prime }_{L^{p}}( \mathbb{R}^{n})$$ onto itself.

Note that if $$\nu \in L^{p}(\mathbb{R}^{n})$$ ($$1< p<\infty$$), then we have
\begin{aligned} \bigl\langle (H\nu)^{\wedge },\varphi \bigr\rangle =& \langle H \nu,\hat{ \varphi } \rangle \\ =& (-1)^{n} \langle \nu,H\hat{\varphi } \rangle \\ =&(-1)^{n} \bigl\langle \check{\nu } ,(H\hat{\varphi })^{\wedge } \bigr\rangle \\ =& (-1)^{n} \bigl\langle \check{\nu } ,(-i)^{n} \operatorname{sgn}(\cdot) \hat{\hat{\varphi }} \bigr\rangle \\ =& \bigl\langle \check{\nu } ,(i)^{n} \operatorname{sgn}(\cdot)\tilde{\varphi } \bigr\rangle \\ =& \bigl\langle \tilde{\check{\nu }} ,(i)^{n} \operatorname{sgn}(\cdot)\varphi \bigr\rangle \\ =& \bigl\langle (-i)^{n} \operatorname{sgn}(\cdot)\hat{\nu } ,\varphi \bigr\rangle \end{aligned}
for all $$\varphi \in \mathcal{S}(\mathbb{R}^{n})$$.
So the following inequality holds:
$$(H\nu)^{\wedge }(x)=(-i)^{n} \operatorname{sgn}(\cdot)\hat{\nu }(x)$$
in the distributional sense.
Let Ω be a nonempty subset of $$\mathbb{R}$$, define (see [16])
$$t\Omega =\{tx:x\in \Omega \},$$
where t is a nonzero real number. Hence we have
$$\operatorname{supp}\biggl(u\biggl(\frac{x}{t}\biggr)\biggr)=t\operatorname{supp}(u).$$
For a subset $$A \subseteq \mathbb{R}$$, define
$$A\Omega =\bigcup_{t\in A}t\Omega.$$

## 2 Main lemmas

In this section, we shall introduce some lemmas.

### Lemma 2.1

Let $$h \in L^{p}(\mathbb{R}^{n})$$ ($$1\leq p<\infty$$) and $$g\in \mathcal{S}(\mathbb{R}^{n})$$. Then the Radon transform of function hg satisfies the Poisson inequality $$\mathfrak{PI}(hg) \leq h \mathfrak{PI}g$$ if and only if
$$p.v. \int_{\mathbb{R}^{n}} \frac{h(x)-h(y)}{\prod^{n}_{j=1}(x_{j}-y_{j})}g(y)\,dy=0,$$
where $$x\in \mathbb{R}^{n}$$.

### Proof

We have
$$\mathfrak{PI}(hg) (x)=\frac{1}{(\pi)^{n}}p.v. \int_{\mathbb{R}^{n}}\frac{h(y)g(y)}{ \prod^{n}_{j=1}(x_{j}-y_{j})}\,dy$$
and
$$h(x)\mathfrak{PI}g(x)=\frac{1}{(\pi)^{n}}p.v. \int_{\mathbb{R}^{n}}\frac{h(x)g(y)}{ \prod^{n}_{j=1}(x_{j}-y_{j})}\,dy$$
for $$x \in \mathbb{R}^{n}$$ from the total Radon transform.
It is clear that the Poisson inequality is satisfied if and only if
$$p.v. \int_{\mathbb{R}^{n}} \frac{h(x)g(y)}{\prod^{n}_{j=1}(x_{j}-y_{j})}\,dy=p.v. \int_{\mathbb{R} ^{n}}\frac{h(y)g(y)}{\prod^{n}_{j=1}(x_{j}-y_{j})}\,dy.$$
So
$$p.v. \int_{\mathbb{R}^{n}} \frac{h(x)-h(y)}{\prod^{n}_{j=1}(x_{j}-y_{j})}g(y)\,dy=0,$$
where $$x\in \mathbb{R}^{n}$$. □
We use $$W^{k,p}(\mathbb{R})$$ to denote the Sobolev space
$$W^{k,p}(\mathbb{R})=\bigl\{ f\in L^{p}(\mathbb{R}):D^{m}f \in L^{p}( \mathbb{R}),\vert m\vert \le k\bigr\} ,$$
where the derivative $$D^{m}f$$ is understood in the distributional sense.

### Lemma 2.2

Suppose that $$1< p\le 2$$. Then, for fixed $$x\in \mathbb{R}$$, the function
$$\nu_{x}(y)=\frac{\mu (x)-\mu (y)}{x-y}$$
for any $$y\in \mathbb{R}$$ and $$\mu \in W^{1,p}(\mathbb{R})$$ is in $$L^{p}(\mathbb{R})$$ and
$$\hat{\nu }(w)=ie^{-ixw} \int^{1}_{0} \frac{w}{t^{2}}e^{\frac{ixw}{t}} \hat{\mu }\biggl(\frac{w}{t}\biggr)\,dt.$$

### Proof

Since $$\mu \in W^{1,p}(\mathbb{R})$$, we have
$$\nu_{x}(y)= \int^{1}_{0}\mu^{\prime }\bigl(ty+(1-t)x\bigr) \,dt.$$
Now we prove that $$\nu \in L^{p}(\mathbb{R})$$. We observe that
\begin{aligned} \Vert \nu \Vert _{p} =& \biggl( \int_{\mathbb{R}}\biggl\Vert \int^{1}_{0}\mu^{\prime }\bigl(ty+(1-t)x\bigr) \,dt \biggr\Vert ^{p} \biggr) ^{\frac{1}{p}} \\ \leq & \int^{1}_{0} \biggl( \int_{\mathbb{R}}\bigl\Vert \mu^{\prime }\bigl(ty+(1-t)x\bigr) \bigr\Vert ^{p}\,dy \biggr) ^{\frac{1}{p}}\,dt \\ =&\bigl\Vert \mu^{\prime }\bigr\Vert _{p} \int^{1}_{0}\frac{1}{\sqrt[p]{t}}\,dt \\ =&p^{\prime }\bigl\Vert \mu^{\prime }\bigr\Vert _{p} \\ < & \infty \end{aligned}
for fixed $$x \in \mathbb{R}$$ by using the generalized Minkowski inequality, which involves that $$\nu \in L^{p}(\mathbb{R})$$.
Since (see [9])
$$\nu =\mathfrak{PI}(u)= \int_{1/\sqrt{k\sigma }}^{u}\sigma (s)\,ds,$$
it follows that
$$\nabla \nu =\sigma (u)\nabla u=\bigl(ku^{2}-1\bigr)^{1/2} \nabla u,$$
which yields that
$$\nabla u=\bigl(ku^{2}-1\bigr)^{-1/2}\nabla \nu.$$
Thus we have (see [11, 22])
$$\bigl(1-ku^{2}\bigr)\nabla u\nabla \varphi =- \bigl(ku^{2}-1\bigr)^{1/2}\nabla \nu \nabla \varphi$$
(2.1)
for each $$\varphi \in C_{0}^{1}(\mathbb{R}^{n})$$.
On the other hand, we have
\begin{aligned} &\int_{\mathbb{R}^{n}}\bigl(ku^{2}-1\bigr)^{1/2}\nabla \nu \nabla \varphi \\ &\quad = \int_{\mathbb{R}^{n}}\nabla \nu \nabla \bigl\{ \bigl(ku^{2}-1 \bigr)^{1/2} \varphi \bigr\} - \int_{\mathbb{R}^{n}}\frac{ku}{ku^{2}-1}\vert \nabla \nu \vert ^{2}\varphi \\ &\quad = - \int_{\mathbb{R^{N}}}a(x)\frac{g(u)}{\sigma (u)}\bigl(ku^{2}-1 \bigr)^{1/2} \varphi - \int_{\mathbb{R^{N}}}ku\vert \nabla u\vert ^{2}\varphi \\ &\quad = - \int_{\mathbb{R^{N}}}a(x)g(u)\varphi - \int_{\mathbb{R^{N}}}ku\vert \nabla u\vert ^{2}\varphi. \end{aligned}
So
\begin{aligned} \hat{\nu }(w) =& \int^{1}_{0} \bigl[ \mu^{\prime }\bigl(ty+(1-t)x \bigr) \bigr] ^{\wedge }(w)\,dt \\ =& e^{-ix\nu } \int^{1}_{0} \frac{1}{t}e^{\frac{ixw}{t}}\hat{ \mu } ^{\prime }\biggl(\frac{w}{t}\biggr)\,dt \\ =&ie^{-ix\nu } \int^{1}_{0} \frac{\nu }{t^{2}}e^{\frac{ixw}{t}} \hat{\mu }\biggl(\frac{w}{t}\biggr)\,dt \end{aligned}
from the definition of $$W^{1,p}(\mathbb{R})$$, which is the desired result. □

## 3 Poisson inequality for $$W^{1,p}(\mathbb{R})$$ functions

In this section, we develop a characterization of $$W^{1,p}(\mathbb{R})$$ functions which satisfy the Poisson inequality $$\mathfrak{PI}(hg) \leq h\mathfrak{PI}g$$.

### Theorem 3.1

Let $$h\in W^{1,p}(\mathbb{R})$$ ($$1< p\le 2$$) and $$g\in L^{p}(\mathbb{R}) \cap L^{p^{\prime }}(\mathbb{R})$$. Then the Radon transform of the function hg satisfies the Poisson inequality $$\mathfrak{PI}(hg) \leq h\mathfrak{PI}g$$ if and only if
$$\int^{1}_{0} \int_{\mathbb{R}}\frac{w}{t^{2}}e^{\frac{-iwx(t-1)}{t}} \hat{h}\biggl( \frac{w}{t}\biggr)\hat{g}(-w)\,dw\,dt=0$$
(3.1)
holds.

### Proof

By Lemma 2.1, we know that $$\mathfrak{PI}hg \leq h\mathfrak{PI}g$$ holds if and only if
$$p.v. \int_{\mathbb{R}^{n}}\frac{h(x)-h(y)}{x-y}g(y)\,dy=0.$$
(3.2)
Since $$h\in W^{1,p}(\mathbb{R})$$, Lemma 2.2 ensures that
$$\frac{h(x)-h(\cdot)}{x-.}\in L^{p}(\mathbb{R}).$$
Thus (3.2) holds if and only if
$$\int_{\mathbb{R}^{n}}\biggl(\frac{h(x)-h(y)}{x-y}\biggr)^{\wedge }(w) \bigl(g(y)\bigr)^{ \vee }(w)\,dw=0,$$
which yields that $$\check{g}(w)=\hat{g}(-w)$$. It is known that the above equation is equivalent to
$$\int_{\mathbb{R}^{n}}ie^{-iwx} \int^{1}_{0}\frac{w}{t^{2}}e^{ \frac{iwx}{t}} \hat{h}\biggl(\frac{w}{t}\biggr)\,dt\hat{g}(-w)\,dw=0$$
from Lemma 2.2.
Let
$$h(t,w)=\frac{w}{t^{2}}e^{\frac{(iwx)(t-1)}{t}}\hat{h}\biggl(\frac{w}{t}\biggr) \hat{g}(-w).$$
Replacing t by $$\frac{1}{y}$$, we obtain that (see [14])
\begin{aligned} \int_{\mathbb{R}} \int^{1}_{0}\bigl\vert h(t,w)\bigr\vert \,dt\,dw =& \int_{\mathbb{R}} \int^{\infty }_{1}\bigl\vert w\hat{h}(wy)\hat{g}(-w) \bigr\vert \,dy\,dw \\ \leq & \biggl( \int_{\mathbb{R}} \int^{\infty }_{1}\bigl\vert y^{-\frac{1+\delta}{p^{\prime }}}\hat{g}(-w) \bigr\vert ^{p}\,dy\,dw \biggr) ^{\frac{1}{p}} \\ &{}\times \biggl( \int_{\mathbb{R}} \int^{\infty }_{1}\bigl\vert wy^{\frac{1+\delta }{p^{\prime }}} \hat{h}(yw) \bigr\vert ^{p^{\prime }}\,dy\,dw \biggr) ^{\frac{1}{p ^{\prime }}} \\ =& \biggl( \frac{p^{\prime }-1}{-p^{\prime }+\delta +2} \biggr) ^{ \frac{1}{p}} \Vert \hat{g} \Vert _{p} \\ &{}\times \biggl( \int_{\mathbb{R}} \int^{\infty }_{1}\bigl\vert wy^{\frac{1+\delta }{p^{\prime }}} \hat{h}(yw) \bigr\vert ^{p^{\prime }}\,dy\,dw \biggr) ^{\frac{1}{p ^{\prime }}} \\ \leq & \biggl( \frac{p^{\prime }-1}{-p^{\prime }+\delta +2} \biggr) ^{ \frac{1}{p}} \Vert \hat{g} \Vert _{p} \\ &{}\times \biggl( \int_{\mathbb{R}} \int^{\infty }_{1}\bigl\vert \lambda \hat{h}(\lambda) \bigr\vert ^{p^{\prime }}y^{\delta -p^{\prime }}\,dy\,d\lambda \biggr) ^{\frac{1}{p^{\prime }}} \\ \leq & \biggl( \frac{p^{\prime }-1}{-p^{\prime }+\delta +2} \biggr) ^{\frac{1}{p}} \Vert \hat{g} \Vert _{p} \\ &{}\times \bigl\Vert \bigl(f^{\prime }\bigr)^{\wedge }\bigr\Vert _{p^{\prime }} \biggl( \frac{1}{p ^{\prime }-\delta -1} \biggr) ^{\frac{1}{p^{\prime }}} \\ \leq & \biggl( \frac{p^{\prime }-1}{-p^{\prime }+\delta +2} \biggr) ^{\frac{1}{p}} \biggl( \frac{1}{p^{\prime }-\delta -1} \biggr) ^{\frac{1}{p ^{\prime }}}\Vert \hat{g} \Vert _{p} \\ &{}\times \bigl\Vert \bigl(f^{\prime }\bigr)\bigr\Vert _{p} \\ < & \infty, \end{aligned}
where
$$\frac{p^{\prime }}{p}-1< \delta < p'-1.$$
Set (see [13])
\begin{aligned}& \Delta w_{\delta }=\overline{a}\bigl(\vert x\vert \bigr) \frac{g(\mathfrak{PI}^{-1}(w_{ \delta }))}{h(\mathfrak{PI}^{-1}(w_{\delta }))} \quad \text{in } \mathbb{R}^{n}, \\& w_{\delta }(0)=\delta, \\& \lim_{\vert x\vert \to \infty } w_{\delta }(x)= \infty, \end{aligned}
and
\begin{aligned}& \Delta w_{\zeta }=\underline{a}\bigl(\vert x\vert \bigr) \frac{g(\mathfrak{PI}^{-1}(w_{ \zeta }))}{h(\mathfrak{PI}^{-1}(w_{\zeta }))} \quad \text{in } \mathbb{R}^{n}, \\& w_{\zeta }(0)=\zeta, \\& \lim_{\vert x\vert \to \infty } w_{\zeta }(x)= \infty, \end{aligned}
respectively.
It follows that
\begin{aligned} w_{\delta }(r) &\leq 2 \int_{0}^{r} \biggl( \int_{0}^{t} \overline{a}(s) \frac{g( \mathfrak{PI}^{-1}(w_{\delta }))}{h(\mathfrak{PI}^{-1}(w_{\delta }))}\,ds \biggr)\,dt \\ &\leq 2g\bigl(\mathfrak{PI}^{-1}\bigl(w_{\delta }(r)\bigr)\bigr) \int_{0}^{r} \biggl( \int _{0}^{t} \frac{\overline{a}(s)}{h(\mathfrak{PI}^{-1}(w_{\delta }))}\,ds \biggr)\,dt \\ &\leq 2g \biggl(\sqrt{2\sqrt{\frac{\varrho }{(\varrho -1)k}} {w_{ \delta }(r)} + \frac{\varrho }{k}} \biggr) \int_{0}^{r} \biggl( \int_{0}^{t} \frac{ \overline{a}(s)}{h(\mathfrak{PI}^{-1} (w_{\delta }))}\,ds \biggr)\,dt \\ &\leq 2g \biggl(2\sqrt[4]{\frac{\varrho }{(\varrho -1)k}} \sqrt{w_{ \delta }} \biggr) \int_{0}^{r} \biggl( \int_{0}^{t} \frac{\overline{a}(s)}{h( \mathfrak{PI}^{-1}(w_{\delta }))}\,ds \biggr)\,dt \\ &\leq \frac{2}{\sqrt{\varrho -1}}g \biggl(2\sqrt[4]{\frac{\varrho }{( \varrho -1)k}} \sqrt{w_{\delta }} \biggr) \biggl[r \biggl( \int_{0}^{r} \overline{a}(t)\,dt \biggr) - \int_{0}^{r}t\overline{a}(t)\,dt \biggr] \\ &\leq \frac{2}{\sqrt{\varrho -1}}g \biggl(2\sqrt[4]{\frac{\varrho }{( \varrho -1)k}} \sqrt{w_{\delta }} \biggr)r \int_{0}^{r}\overline{a}(t)\,dt \end{aligned}
for all $$r>0$$ sufficiently large, which yields that (see [17])
$$2\sqrt[4]{\frac{\varrho }{(\varrho -1)k}}\sqrt{w_{\delta }} \leq \mathcal{G}^{-1} \biggl(r \int_{0}^{r} \overline{a}(t)\,dt \biggr)$$
for all $$r\gg0$$.
Put
$$0< S(\zeta)=\sup \bigl\{ r>0 :w_{\delta }(r)< w_{\zeta }(r)\bigr\} \leq \infty.$$
So
\begin{aligned} \zeta_{0} \leq &\delta + \int_{0}^{S(\zeta_{0})}t^{1-N} \biggl[ \int _{0}^{t}s^{N-1} \biggl( \overline{a}(s)\frac{g(\mathfrak{PI}^{-1}(w_{ \delta }))}{h(\mathfrak{PI}^{-1}(w_{\delta }))} -\underline{a}(s)\frac{g( \mathfrak{PI}^{-1}(w_{\zeta }))}{h(\mathfrak{PI}^{-1}(w_{\zeta }))} \biggr)\,ds \biggr]\,dt \\ \leq& \delta + \int_{0}^{S(\zeta_{0})}t^{1-N} \biggl[ \int _{0}^{t}s^{N-1} \biggl( \overline{a}(s)\frac{g(\mathfrak{PI}^{-1}(w_{ \delta }))}{h(\mathfrak{PI}^{-1}(w_{\delta }))} \\ &{} - \underline{a}(s)\frac{g(\mathfrak{PI}^{-1}(w_{\zeta }))}{ \mathfrak{PI}^{-1}(w_{\zeta })^{\delta }} \frac{\mathfrak{PI}^{-1}(w _{\zeta })^{\delta }}{h(\mathfrak{PI}^{-1}(w_{\zeta }))} \biggr)\,ds \biggr] \,dt \\ \leq& \delta + \int_{0}^{S(\zeta_{0})}t^{1-N} \biggl[ \int_{0}^{t}s^{N-1} \biggl(\overline{a}(s) \frac{g(\mathfrak{PI}^{-1}(w_{\delta }))}{h( \mathfrak{PI}^{-1}(w_{\delta }))} -\underline{a}(s)\frac{g( \mathfrak{PI}^{-1}(w_{\delta }))}{h(\mathfrak{PI}^{-1}(w_{\delta }))} \biggr)\,ds \biggr]\,dt. \end{aligned}
(3.3)
On the other hand, we have
\begin{aligned} 0 &\leq t^{1-N} \biggl[ \int_{0}^{t}s^{N-1} \biggl(\overline{a}(s) \frac{g( \mathfrak{PI}^{-1}(w_{\delta }))}{h(\mathfrak{PI}^{-1}(w_{\delta }))} -\underline{a}(s)\frac{g(\mathfrak{PI}^{-1}(w_{\delta }))}{h( \mathfrak{PI}^{-1}(w_{\delta }))} \biggr)\,ds \biggr] \chi_{[0,S(\zeta)]}(t) \\ &= t^{1-N} \biggl[ \int_{0}^{t}s^{N-1}a_{\mathrm{osc}}(s) \frac{g( \mathfrak{PI}^{-1}(w_{\delta }))}{h(\mathfrak{PI}^{-1}(w_{\delta }))}\,ds \biggr] \\ &\leq \frac{1}{\sqrt{\varrho -1}} \biggl(t^{1-N} \int_{0}^{t}s^{N-1}a _{\mathrm{osc}}(s) \,ds \biggr) g \biggl(\mathcal{G}^{-1} \biggl(t \int_{0}^{t} \overline{a}(s)\,ds \biggr) \biggr): = \mathcal{H}(t) \end{aligned}
for $$t\gg 0$$, where $$\chi_{[0,S(\zeta)]}$$ stands for the characteristic function of $$[0,S(\zeta)]$$, which yields that (see [12])
$$\zeta_{0} \leq \delta + \int_{0}^{\infty }\mathcal{H}(s)\,ds\leq \delta + \overline{H},$$
but this is impossible.
Consider the following problem (see [15]):
\begin{aligned} &\Delta w={a}(x) \frac{g(\mathfrak{PI}^{-1}(w))}{h(\mathfrak{PI}^{-1}(w))} \quad \text{in } B_{n}(0), \\ &w\geq 0\quad \text{ in }B_{n}(0), \\ &w=w_{\delta }\quad \text{on } \partial B_{n}(0). \end{aligned}
(3.4)
As a consequence, we get
$$\int^{1}_{0} \int_{\mathbb{R}}\frac{w}{t^{2}}e^{\frac{-iwx(t-1)}{t}} \hat{h}\biggl( \frac{w}{t}\biggr)\hat{g}(-w)\,dw\,dt=0$$
by using the Fubini theorem.

This completes the proof. □

Now we give an application of Theorem 3.1.

### Theorem 3.2

Let $$h\in W^{1,p}(\mathbb{R})$$ ($$1< p\le 2$$) and $$g\in \mathfrak{PI}^{p}( \mathbb{R})\cap \mathfrak{PI}^{p^{\prime }}(\mathbb{R})$$. If
$$\bigl(I^{-}\operatorname{supp}\hat{h}\bigr)\cap \operatorname{supp}\hat{g}=\emptyset,$$
(3.5)
where $$I^{-}=[-1,0)$$, then the Poisson inequality $$\mathfrak{PI}(hg) \leq h\mathfrak{PI}g$$ holds.

### Proof

By condition (3.5), we obtain that (see [4])
$$(t\operatorname{supp}\hat{h})\cap \operatorname{supp}\hat{g}=\emptyset$$
for any $$t \in I^{-}$$, which is equivalent to
$$\operatorname{supp}\hat{h}\biggl(\frac{.}{t}\biggr)\cap \operatorname{supp}\hat{g}= \emptyset$$
for any $$t \in I^{-}$$.
By the embedding theorem and Hölder’s inequality, we obtain
\begin{aligned}[b] & \int_{A_{k_{j+1},j+1}}\bigl(h(u)-k_{j+1}\bigr)_{+}\,dx \\ &\quad \leq \biggl( \int_{A_{k_{j+1},j}}\bigl(\bigl(h(u)-k_{j+1}\bigr)_{+} \zeta_{j}^{q}\bigr)^{\frac{n}{n-1}}\,dx \biggr)^{\frac{n-1}{n}} \vert A_{k_{j+1},j+1}\vert ^{1/n} \\ &\quad \leq \gamma \int_{A_{k_{j+1},j}}\bigl\vert \nabla \bigl(\bigl(h(u)-k_{j+1} \bigr)_{+}\zeta_{j}^{q}\bigr)\bigr\vert \vert A_{k_{j+1},j}\vert ^{1/n} \\ &\quad \leq \gamma \biggl( \int_{A_{k_{j+1},j}}g(u)\vert \nabla u\vert \zeta_{j}^{q} \,dx \\ &\qquad {} + \int_{A_{k_{j+1},j}}\bigl(h(u)-k_{j+1}\bigr)_{+} \vert \nabla \zeta_{j}\vert \zeta_{j}^{q-1}\,dx \biggr) \vert A_{k_{j+1},j}\vert ^{1/n}. \end{aligned}
(3.6)
Let $$\ell =\delta (\rho)/\rho$$. We estimate the first term on the right-hand side of (3.6) as follows:
\begin{aligned}[b] & \int_{A_{k_{j+1},j}}g(u)\vert \nabla u\vert \zeta_{j}^{q} \,dx \\ &\quad =\frac{1}{g( \ell)} \int_{A_{k_{j+1},j}}g(u)g(\ell)\vert \nabla u\vert \zeta_{j}^{q}\,dx \\ & \quad \leq \ell \int_{A_{k_{j+1},j}}g(u)\zeta_{j}^{q}\,dx + \frac{1}{g(\ell)} \int_{A_{k_{j+1},j}}g(u)G\bigl(\vert \nabla u\vert \bigr) \zeta_{j}^{q}\,dx \\ &\quad \leq 2^{j}\frac{ \ell }{k} \int_{A_{k_{j+1},j}}\bigl(h(u)-k_{j+1}\bigr)_{+}g(u) \zeta_{j}^{q}\,dx +\frac{1}{g( \ell)} \int_{A_{k_{j+1},j}}g(u)G\bigl(\vert \nabla u\vert \bigr) \zeta_{j}^{q}\,dx. \end{aligned}
(3.7)
It follows that
\begin{aligned}[b] & \int_{A_{k_{j+1},j}}g(u)\vert \nabla u\vert \zeta_{j}^{q} \,dx \\ &\quad \leq \gamma (1- \varrho)^{-\gamma }2^{j\gamma } \biggl( \frac{\ell }{k}+\frac{1}{g( \ell)} \biggr)\rho^{-1} g \biggl( \frac{\delta (\rho)}{\rho } \biggr) \int_{A_{k_{j},j}}\bigl(h(u)-k_{j}\bigr)_{+}\,dx \end{aligned}
(3.8)
from the previous inequality and Lemma 2.2.
Since
$$k\geq G(\ell)=G \biggl( \frac{\delta (\rho)}{\rho } \biggr),$$
(3.9)
we obtain
$$y_{j+1}= \int_{A_{k_{j+1},j+1}}\bigl(h(u)-k_{j+1}\bigr)\,dx \leq \gamma (1- \varrho)^{-\gamma }2^{j\gamma }\rho^{-1} k^{-\frac{1}{n}}y_{j}^{1+ \frac{1}{n}}$$
(3.10)
from (3.6) and (3.7), which gives that
$$k\geq \gamma (1-\varrho)^{-\gamma }\rho^{-n} \int_{B_{\frac{1-\varrho }{2}\rho }(\bar{x})}h(u)\,dx.$$
(3.11)
(3.8) and (3.9) also imply that
$$h\bigl(u(\bar{x})\bigr)\leq \gamma (1-\varrho)^{-\gamma }G \biggl( \frac{\delta ( \rho)}{\rho } \biggr) + \gamma (1-\varrho)^{-\gamma } \rho^{-n} \int_{B_{\frac{1-\varrho }{2}\rho }(\bar{x})} h(u)\,dx.$$
(3.12)
Since
\begin{aligned} \int_{B_{\frac{1-\varrho }{2}\rho }(\bar{x})}h(u)\,dx &\leq \delta ( \rho) \int_{B_{(1-\varrho)\rho }(\bar{x})}g(u)\xi^{q}\,dx \\ &\leq \gamma (1-\varrho)^{-1}\frac{\delta (\rho)}{\rho } \int_{B_{(1-\varrho)\rho }(\bar{x})}g\bigl(\vert \nabla u\vert \bigr) \xi^{q-1}\,dx, \end{aligned}
we obtain that
\begin{aligned}[b] \int_{B_{\frac{1-\varrho }{2}\rho }(\bar{x})}h(u)\,dx\leq{}& \gamma (1- \varrho)^{-1} \frac{\delta (\rho)}{M(\rho)} \int_{B_{(1-\varrho)\rho }(\bar{x})}G\bigl(\vert \nabla u\vert \bigr) \xi^{q}\,dx \\ &{} + \gamma (1-\varrho)^{-1}\frac{\delta (\rho)}{\rho } g \biggl( \frac{M( \rho)}{\rho } \biggr) \rho^{n} \end{aligned}
(3.13)
and
$$\int_{B_{(1-\varrho)\rho }(\bar{x})}G\bigl(\vert \nabla u\vert \bigr) \xi^{q}\,dx \leq \gamma (1-\varrho)^{-\gamma }G \biggl( \frac{M(\rho)}{\rho } \biggr) \rho ^{n}.$$
(3.14)
Combining (3.13) and (3.14), we have
$$\int_{B_{(1-\varrho)\rho }(\bar{x})}h(u)\,dx \leq \gamma (1-\varrho)^{- \gamma } \frac{\delta (\rho)}{\rho } g \biggl( \frac{M(\rho)}{\rho } \biggr) \rho^{n}.$$
(3.15)
As a consequence, (3.1) holds. Thus, by invoking Theorem 3.1, the Radon transform of function hg satisfies the Poisson inequality
$$\mathfrak{PI}(hg)\leq h\mathfrak{PI}g.$$
□

## 4 Conclusions

This paper was mainly devoted to studying a new Poisson inequality for the Radon transform of infinitely differentiable functions. An application of it was also given.

## Declarations

### Acknowledgements

The author is thankful to the honorable reviewers for their valuable suggestions and comments, which improved the paper.

Not applicable.

Not applicable.

### Authors’ contributions

The author read and approved the final manuscript.

### Competing interests

The author declares that he has no competing interests.

## Authors’ Affiliations

(1)
School of Economic Mathematics, Southwestern University of Finance and Economic, Chengdu, China

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