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New Poisson inequality for the Radon transform of infinitely differentiable functions
Journal of Inequalities and Applications volume 2018, Article number: 214 (2018)
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).
1 Introduction
The Radon transform \(\mathfrak{PI}\), which is defined as the Cauchy principal value of the following singular integral
for any \(x\in \mathbb{R}\), has been widely used in physics, engineering, and mathematics. The following Poisson inequality
was first studied in [1–3, 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
The total Radon transform \(\mathfrak{PI}\) of a function \(h \in L^{p}( \mathbb{R}^{n})\) (\(1 \le p< \infty\)) is defined as follows:
The existence of the singular integral above and its boundedness property
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
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]):
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
and
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
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:
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
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
for any \(\varphi \in \mathcal{S}(\mathbb{R}^{n})\), where
and ν̃ is defined as follows:
So we obtain that
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
-
(I)
\(\varphi \in C^{\infty }(\mathbb{R}^{n})\);
-
(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
where
As a consequence, we have
and
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])
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
for all \(\varphi \in \mathcal{S}(\mathbb{R}^{n})\).
So the following inequality holds:
in the distributional sense.
Let Ω be a nonempty subset of \(\mathbb{R}\), define (see [16])
where t is a nonzero real number. Hence we have
For a subset \(A \subseteq \mathbb{R}\), define
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
where \(x\in \mathbb{R}^{n}\).
Proof
We have
and
for \(x \in \mathbb{R}^{n}\) from the total Radon transform.
It is clear that the Poisson inequality is satisfied if and only if
So
where \(x\in \mathbb{R}^{n}\). □
We use \(W^{k,p}(\mathbb{R})\) to denote the Sobolev space
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
for any \(y\in \mathbb{R}\) and \(\mu \in W^{1,p}(\mathbb{R})\) is in \(L^{p}(\mathbb{R})\) and
Proof
Since \(\mu \in W^{1,p}(\mathbb{R})\), we have
Now we prove that \(\nu \in L^{p}(\mathbb{R})\). We observe that
for fixed \(x \in \mathbb{R}\) by using the generalized Minkowski inequality, which involves that \(\nu \in L^{p}(\mathbb{R})\).
Since (see [9])
it follows that
which yields that
for each \(\varphi \in C_{0}^{1}(\mathbb{R}^{n})\).
On the other hand, we have
So
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
holds.
Proof
By Lemma 2.1, we know that \(\mathfrak{PI}hg \leq h\mathfrak{PI}g\) holds if and only if
Since \(h\in W^{1,p}(\mathbb{R})\), Lemma 2.2 ensures that
Thus (3.2) holds if and only if
which yields that \(\check{g}(w)=\hat{g}(-w)\). It is known that the above equation is equivalent to
from Lemma 2.2.
Let
Replacing t by \(\frac{1}{y}\), we obtain that (see [14])
where
Set (see [13])
and
respectively.
It follows that
for all \(r>0\) sufficiently large, which yields that (see [17])
for all \(r\gg0\).
Put
So
On the other hand, we have
for \(t\gg 0\), where \(\chi_{[0,S(\zeta)]}\) stands for the characteristic function of \([0,S(\zeta)]\), which yields that (see [12])
but this is impossible.
Consider the following problem (see [15]):
As a consequence, we get
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
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])
for any \(t \in I^{-}\), which is equivalent to
for any \(t \in I^{-}\).
By the embedding theorem and Hölder’s inequality, we obtain
Let \(\ell =\delta (\rho)/\rho \). We estimate the first term on the right-hand side of (3.6) as follows:
It follows that
from the previous inequality and Lemma 2.2.
Since
we obtain
from (3.6) and (3.7), which gives that
(3.8) and (3.9) also imply that
Since
we obtain that
and
Combining (3.13) and (3.14), we have
As a consequence, (3.1) holds. Thus, by invoking Theorem 3.1, the Radon transform of function hg satisfies the Poisson inequality
□
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.
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Sun, Z. New Poisson inequality for the Radon transform of infinitely differentiable functions. J Inequal Appl 2018, 214 (2018). https://doi.org/10.1186/s13660-018-1805-9
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DOI: https://doi.org/10.1186/s13660-018-1805-9