Sets of $p$-restriction and $p$-spectral synthesis

TL;DR

This paper explores conditions under which sets in Euclidean space fail to have the p-restriction property and extends spectral synthesis concepts to p-restriction sets, demonstrating the sphere's spectral synthesis for certain p-values.

Contribution

It provides new sufficient conditions for the failure of p-restriction and extends spectral synthesis to L^p spaces for sets with p-restriction, including the unit sphere.

Findings

Identifies conditions for failure of p-restriction property.

Extends spectral synthesis to L^p spaces for p > 1.

Shows the unit sphere is p-spectral in certain dimensions.

Abstract

In this paper we investigate the restriction problem. More precisely, we give sufficient conditions for the failure of a set $E$ in $\mathbb{R}^n$ to have the $p$-restriction property. We also extend the concept of spectral synthesis to $L^p(\mathbb{R}^n)$ for sets of $p$-restriction when $p > 1$. We use our results to show that there are $p$-values for which the unit sphere is a set of $p$-spectral synthesis in $\mathbb{R}^n$ when $n \geq 3$.

Equations

$$\hat{f}(\xi )=\mathcal{F}(f)(\xi )={\int }_{{\mathbb{R}}^n}{e}^{-2\pi ix\cdot \xi }f(x)dx,\xi \in {\mathbb{R}}^n.$$

$${\mathcal{R}}_E(f)=\hat{f}{\mid }_E$$

$$J(E)=\{f\in \mathcal{S}({\mathbb{R}}^n)\mid \hat{f}{\mid }_E=0\},$$

$$I^p(E)=\{f\in L^p({\mathbb{R}}^n)\mid \hat{f}{\mid }_E=0\}.$$

$$\left(\frac{{\partial }^2\varphi }{\partial x_i\partial x_j}\right)$$

$$Z(f)=\{\xi \in {\mathbb{R}}^n\mid \hat{f}(\xi )=0\}.$$

$$Z(I)=\bigcap _{f\in I}Z(f).$$

$$k(E)=\{f\in \mathcal{S}({\mathbb{R}}^n)\mid \hat{f}=0\text{mbox}onaneighborhoodofE\}.$$

$$k(E)\subseteq J(E)\subseteq L^p({\mathbb{R}}^n)$$

$${\overline{k(E)}}^p=I^p(E).$$

$$f\ast g(x)={\int }_{{\mathbb{R}}^n}f(x-y)g(y)dy.$$

$$T_{\varphi }(f)={\int }_{{\mathbb{R}}^n}f(-x)\varphi (x)dx.$$

$$\text{Ann}(X)=\{\varphi \in L^{p^{\prime }}({\mathbb{R}}^n)\mid T_{\varphi }(f)=0\text{ for all }f\in X\},$$

$$f\ast \varphi (x)={\int }_{{\mathbb{R}}^n}f(x-y)\varphi (y)dy={\int }_{{\mathbb{R}}^n}{f}_{-x}(-y)\varphi (y)dy=T_{\varphi }(f_{-x}).$$

$$\check{f}(x)={\int }_{{\mathbb{R}}^n}f(\xi )e^{2\pi i(\xi \cdot x)}d\xi$$

$$\hat{T}(f)=T(\hat{f})$$

$$\check{T}(f)=T(\check{f}),$$

$$a_n=n^{-2}[\underset{B}{\sup }(f_{x_n}){]}^{-1}.$$

$$g=\sum _{n=1}^{\infty }{a}_n{f}_{x_n}$$

$$\Vert \hat{f}{\Vert }_{L^1(E)}\le C\Vert f{\Vert }_{L^p}$$

$$\Vert \check{Fd\sigma }{\Vert }_{L^{p^{\prime }}}\le C\Vert F{\Vert }_{L^{\infty }(E)}$$

$$\check{d\mu }(x)={\int }_{{\mathbb{R}}^n}{e}^{2\pi i(x\cdot \xi )}d\mu (\xi ),$$

$$|f\ast \varphi (x)|=\underset{n\to \infty }{\lim }|f_n\ast \varphi (x)|=\underset{n\to \infty }{\lim }\left|{\int }_{{\mathbb{R}}^n}{e}^{2\pi i(x\cdot \xi )}\widehat{f_n}(\xi )d\sigma (\xi )\right|.$$

$$\underset{n\to \infty }{\lim }\left|{\int }_E{e}^{2\pi i(x\cdot \xi )}\widehat{f_n}(\xi )d\sigma (\xi )\right|$$

$$\{\check{Fd\sigma }\mid F\text{ is smooth on }E\}.$$

$$⟨f,\varphi ⟩={\int }_{S^{n-1}}\frac{\partial \hat{f}}{\partial x_n}d\sigma .$$

Full text

1. Introduction

Throughout this paper $\mathbb{R}$ will denote the real numbers and $\mathbb{Z}$ will denote the integers. Let $p\in[1,\infty]$ and let $n$ be a positive integer. Indicate by $L^{p}(\mathbb{R}^{n})$ the usual Lebesgue space and denote by $\|\cdot\|_{L^{p}}$ the usual Banach space norm on $L^{p}(\mathbb{R}^{n})$. If $X\subset L^{p}(X)$, then $\overline{X}^{p}$ will denote the closure of $X$ in $L^{p}(\mathbb{R}^{n})$. Also $p^{\prime}$ will always represent the conjugate index of $p$, that is $\frac{1}{p}+\frac{1}{p^{\prime}}=1$. For $f\in L^{1}(\mathbb{R}^{n})$ the Fourier transform of $f$ is defined by

[TABLE]

The Fourier transform can be extended to a unitary operator on $L^{2}(\mathbb{R}^{n})$ and by the Hausdorff-Young inequality, $\mathcal{F}$ can be extended to a continuous operator from $L^{p}(\mathbb{R}^{n})$ to $L^{p^{\prime}}(\mathbb{R}^{n})$, when $1<p<2$. Let $E$ be a closed subset of $\mathbb{R}^{n}$. If $f\in L^{1}(\mathbb{R}^{n})$, then $\hat{f}$ is continuous on $\mathbb{R}^{n}$. Consequently, the restriction of $\hat{f}$ to $E$, which we denote by $\hat{f}\mid_{E}$, is a well-defined function on $E$. For $1<p<2$ and $f\in L^{p}(\mathbb{R}^{n}),\hat{f}\in L^{p^{\prime}}(\mathbb{R}^{n})$. So if $E$ is a set of positive Lebesgue measure we can restrict $\hat{f}$ to $E$. The interesting question is can $\hat{f}$ be restricted to $E$ when $E$ has Lebesgue measure zero? This question is the heart of the restriction problem, which we will now describe.

For $E\subseteq\mathbb{R}^{n}$, let $C(E)$ be the set of continuous functions on $E$ and let $C_{c}(E)$ be the set of functions in $C(E)$ with compact support. Let $L^{p}(E)$ be the usual Banach space formed with respect to the induced measure $d\sigma$ on $E$. The norm on $L^{p}(E)$ will be denoted by $\|\cdot\|_{L^{p}(E)}$. Recall that the norm on $L^{p}(\mathbb{R}^{n})$ is indicated by $\|\cdot\|_{L^{p}}$. Let $2\leq n\in\mathbb{Z}$ and let $\mathcal{S}(\mathbb{R}^{n})$ denote the space of Schwartz functions on $\mathbb{R}^{n}$. The operator $\mathcal{R}_{E}\colon\mathcal{S}(\mathbb{R}^{n})\rightarrow C(E)$ given by

[TABLE]

is known as the restriction operator associated with $E$. If $\mathcal{R}_{E}$ can be extended to a continuous operator from $L^{p}(\mathbb{R}^{n})\rightarrow L^{q}(E)$, then we shall say that $\mathcal{R}_{E}$ has property $\mathcal{R}(E,p,q)$. Observe that if $1\leq q_{1}\leq q_{2}$ and $\mathcal{R}_{E}$ has property $\mathcal{R}(E,p,q_{2})$, then it also has property $\mathcal{R}(E,p,q_{1})$. We shall say that $E$ is a set of $p$-restriction if $\mathcal{R}_{E}$ has property $\mathcal{R}(E,p,1)$. Note that any closed set in $\mathbb{R}^{n}$ is a set of $1$-restriction. Furthermore, if $E$ is not a set of $p$-restriction, then $\mathcal{R}_{E}$ does not have property $\mathcal{R}(E,p,q)$ for any $q\geq 1$. The best known result concerning the restriction of $\hat{f}$ to $E$ is the Stein-Tomas theorem: $\mathcal{R}_{E}$ has property $\mathcal{R}(E,p,2)$ if and only if $1\leq p\leq\frac{2n+2}{n+3}$, where $E$ is a smooth compact hypersurface in $\mathbb{R}^{n}$ with nonzero Gaussian curvature. In general though it is an extremely difficult problem to determine if $\mathcal{R}_{E}$ has property $\mathcal{R}(E,p,q)$. A more comprehensive treatment of the restriction problem, along with its history, can be found in [3, 11], and [4, Chapter 5.4] and the references therein.

In this paper we will only be concerned with the case where $\mathcal{R}_{E}$ has property $\mathcal{R}(E,p,1)$, that is $E$ is a set of $p$-restriction. Set

[TABLE]

and if $E$ is a set of $p$-restriction define

[TABLE]

This paper was inspired by the paper [10] where these spaces were investigated for the case when $E$ is the unit sphere $S^{n-1}$ in $\mathbb{R}^{n}$. Our first main result is:

Theorem 1.1**.**

Let $E$ be a smooth compact submanifold of codimension $k$ in $\mathbb{R}^{n}$. If there exists $f\in C_{c}(\mathbb{R}^{n})$ such that $\hat{f}$ vanishes on $E$, then $E$ is not a set of $p$-restriction for $p>\frac{2n}{n+k}$.

Thus $\mathcal{R}_{E}$ does not have property $\mathcal{R}(E,p,q)$ when $p>\frac{2n}{n+k}$ and $q\geq 1$. If $E$ is a hypersurface, then the lower bound for $p$ becomes $\frac{2n}{n+1}$. We are able to improve this lower bound for hypersurfaces with the constant relative nullity condition, which we now define. Let $U$ be an open set in $\mathbb{R}^{n-1}$ and let $F=\{(x,\phi(x))\mid x\in U\}$ be a smooth hypersurface in $\mathbb{R}^{n}$. If the Hessian matrix

[TABLE]

of $\phi$ has constant rank $n-1-\nu$ on $U$, where $0\leq\nu\leq n-1$, then we say that $\phi$ has constant relative nullity $\nu$. A smooth hypersurface $E$ of $\mathbb{R}^{n}$ is said to have constant relative nullity $\nu$ if every localization $F$ of $E$ has constant relative nullity $\nu$. If $\nu=n-1$, then $E$ is a hyperplane. It is known that hyperplanes are sets of $p$-restriction only if $p=1$. Thus we will only consider hypersurfaces with $0\leq\nu\leq n-2$. Note that $\nu=0$ for $S^{n-1}$.

Theorem 1.2**.**

Let $2\leq n\in\mathbb{Z}$ and let $E$ be a smooth compact hypersurface in $\mathbb{R}^{n}$ with constant relative nullity $\nu$, for $0\leq\nu\leq n-2$. If $\frac{2(n-\nu)}{n-\nu+1}<p\in\mathbb{R}$, then $E$ is not a set of $p$-restriction.

Let $f\in L^{p}(\mathbb{R}^{n})$ and let $y\in\mathbb{R}^{n}$. The translate of $f$ by $y$, which we write as $f_{y}$, is the function $f_{y}(x)=f(x-y)$, where $x\in\mathbb{R}^{n}$. For $f\in L^{p}(\mathbb{R}^{n})$, let $T^{p}[f]$ be the closed subspace of $L^{p}(\mathbb{R}^{n})$ spanned by $f$ and its translates. The zero set $Z(f)$ of $f\in L^{1}(\mathbb{R}^{n})$ is defined by

[TABLE]

In Section 3 we will see that if $Z(f)$ is a set of $p$-restriction, then $T^{p}[f]\neq L^{p}(\mathbb{R}^{n})$.

We will now briefly review the concept of spectral synthesis in $L^{1}(\mathbb{R}^{n})$. Suppose $I$ is a closed ideal in $L^{1}(\mathbb{R}^{n})$ and define the zero set of $I$ by

[TABLE]

Let $E$ be a closed set in $\mathbb{R}^{n}$, then $I^{1}(E)$ is a closed ideal in $L^{1}(\mathbb{R}^{n})$ with zero set $E$. In fact, $I^{1}(E)$ is the largest closed ideal in $L^{1}(\mathbb{R}^{n})$ whose zero set is $E$. Now let

[TABLE]

Then

[TABLE]

and $\overline{k(E)}^{1}$ is the smallest closed ideal in $L^{1}(\mathbb{R}^{n})$ with zero set $E$. The set $E$ is known as a set of spectral synthesis if $\overline{k(E)}^{1}=I^{1}(E)$. A more detailed account of spectral synthesis can be found in [2][9, Chapter 7]. Extending the concept of spectral synthesis to $L^{p}(\mathbb{R}^{n})$ for $p>1$ falls short since the analog to $I^{1}(E),I^{p}(E)$, is not well-defined for closed sets of Lebesgue measure zero in $\mathbb{R}^{n}$. However, for sets of $p$-restriction $I^{p}(E)$ is well-defined, which allows us to extend the idea of spectral synthesis to $L^{p}(\mathbb{R}^{n})$ for sets $E$ of $p$-restriction. We shall say that a set $E$ of $p$-restriction is a set of $p$-spectral synthesis if

[TABLE]

We can now state:

Theorem 1.3**.**

Let $2\leq n\in\mathbb{Z}$ and let $E$ be a smooth compact hypersurface in $\mathbb{R}^{n}$ with constant relative nullity $\nu,0\leq\nu\leq n-2$. If $E$ is a set of $p$-restriction for some $p$ that satisfies one of the following:

1. (1)

$\frac{2(n-\nu)}{n+3-\nu}\leq p<2$* and $0\leq\nu<n-3$* 2. (2)

$1<p<2$* for $n-3\leq\nu<n-1$,*

then $E$ is a set of $p$-spectral synthesis.

It is known that $S^{1}$ is a set of spectral synthesis in $\mathbb{R}^{2}$ [8], but $S^{n-1}$ is not a set of spectral synthesis in $\mathbb{R}^{n}$ for $n\geq 3$ [9, Chapter 7.3]. We will use Theorem 1.3 to show that there are $p$-values where $S^{n-1}$ is a set of $p$-spectral synthesis for $n\geq 3$.

This paper is organized as follows: In Section 2 we give some background and results that will be needed for this paper. In Section 3 we will prove Theorems 1.1 and 1.2 by linking them to the problem of determining when $T^{p}[f]$ is dense in $L^{p}(\mathbb{R}^{n})$ for $f\in\mathcal{S}(\mathbb{R}^{n})$ with $\hat{f}=0$ on $E$. In Section 4 we prove Theorem 1.3, and use the theorem to show that there are $p$-values for which the unit sphere $S^{n-1}$ is a set of $p$-spectral synthesis in $\mathbb{R}^{n}$ for $n\geq 3$.

2. Preliminaries

In this section we will give some results that will be used in the sequel. The convolution of two measurable functions $f$ and $g$ on $\mathbb{R}^{n}$ is defined by

[TABLE]

Let $1<p<2$. Each $\phi\in L^{p^{\prime}}(\mathbb{R}^{n})$ defines a bounded linear functional $T_{\phi}$ on $L^{p}(\mathbb{R}^{n})$ via

[TABLE]

Sometimes we will write $\langle f,\phi\rangle$ in place of $T_{\phi}(f)$. For closed subspaces $X$ in $L^{p}(\mathbb{R}^{n})$,

[TABLE]

will denote the annihilator of $X$ in $L^{p^{\prime}}(\mathbb{R}^{n})$. The following characterization of $\text{Ann}(X)$ when $X$ is a translation-invariant subspace of $L^{p}(\mathbb{R}^{n})$ will be needed later.

Proposition 2.1**.**

Let $X$ be a translation-invariant subspace of $L^{p}(\mathbb{R}^{n})$. Then $\phi\in\text{Ann}(X)$ if and only if $f\ast\phi=0$ for all $f\in X$.

Proof.

Observe that for $f\in L^{p}(\mathbb{R}^{n})$ and $\phi\in L^{p^{\prime}}(\mathbb{R}^{n})$

[TABLE]

It follows from the translation invariance of $X$ that $f\ast\phi=0$ for all $f\in X$ if and only if $\phi\in\text{Ann}(X)$. ∎

The space $L^{p}(\mathbb{R}^{n})$ is a $L^{1}(\mathbb{R}^{n})$-module since $f\ast g\in L^{p}(\mathbb{R}^{n})$ whenever $f\in L^{1}(\mathbb{R}^{n})$ and $g\in L^{p}(\mathbb{R}^{n})$. The following proposition will not be used in the paper, but we record it here for its independent interest.

Proposition 2.2**.**

If $E$ is a set of $p$-restriction, then $I^{p}(E)$ is a $L^{1}(\mathbb{R}^{n})$-submodule of $L^{p}(\mathbb{R}^{n})$.

Proof.

A modification of the proof of [9, Theorem 7.1.2] will show that a closed translation-invariant subspace of $L^{p}(\mathbb{R}^{n})$ is translation invariant if and only if it is a $L^{1}(\mathbb{R}^{n})$-submodule of $L^{p}(\mathbb{R}^{n})$. The proposition now follows since $I^{p}(E)$ is a closed translation-invariant subspace of $L^{p}(\mathbb{R}^{n})$. ∎

It is well known that the Fourier transform is an isomorphism on the Schwartz space $\mathcal{S}(\mathbb{R}^{n})$, with inverse Fourier transform given by

[TABLE]

for $f\in\mathcal{S}(\mathbb{R}^{n})$. A continuous linear functional on $\mathcal{S}(\mathbb{R}^{n})$ is known as a temperate distribution. A nice property of temperate distributions is that the Fourier transform can be extended to them. In fact, the Fourier transform defines an isomorphism on the temperate distributions. Indeed, if $T$ is a tempered distribution, then $\widehat{T}$ is the tempered distribution given by

[TABLE]

for $f\in\mathcal{S}(\mathbb{R}^{n})$. The inverse Fourier transform $\widecheck{T}$ of a temperate distribution $T$ is defined by

[TABLE]

where $f\in\mathcal{S}(\mathbb{R}^{n})$. Since elements of $L^{p}(\mathbb{R}^{n})$ are temperate distributions, we can define the Fourier transform $\widehat{f}$ for $f\in L^{p}(\mathbb{R}^{n})$ in the distributional sense when $p>2$. For the rest of this paper, distribution will mean temperate distribution.

We shall write $\text{supp}(\psi)$ to indicate the support of $\psi$, where depending on the context, $\psi$ is a function, measure, or distribution.

We conclude this section with a result that will be needed later.

Proposition 2.3**.**

If $E$ is a compact subset of $\mathbb{R}^{n}$, then there exists an $f\in\mathcal{S}(\mathbb{R}^{n})$ for which $Z(f)=E$.

Proof.

Let $B$ be an open ball containing $E$ and let $x\in B\setminus E$. The Whitney extension theorem produces a smooth function $f_{x}\colon B\rightarrow\mathbb{R}$ such that $f_{x}=0$ on $E$ and $f_{x}>0$ at $x$. For the purpose of this proof only, $f_{x}$ will mean the function defined above instead of the translate of $f$. For each $x\in B\setminus E$ there exists an open ball $B_{x}$ for which $f_{x}$ is positive on $B_{x}$. Now choose a countable subcover $B_{x_{n}}$ of $B\setminus E$. Let

[TABLE]

Then

[TABLE]

is a smooth function on $B$. Let $B_{1}$ be an open ball satisfying $E\subseteq B_{1}\subseteq B.$ Deonte by $h$ the smooth function obtained by multiplying $g$ by a smooth function that equals one on $B_{1}$ and zero on $\mathbb{R}^{n}\setminus B$. Set $F=h+s$ where $s\in\mathcal{S}(\mathbb{R}^{n})$ that is zero on $B_{1}$ and positive on $\mathbb{R}^{n}\setminus\overline{B_{1}}$, where $\overline{B_{1}}$ is the closure of $B_{1}$. Thus $F\in\mathcal{S}(\mathbb{R}^{n})$ and can be expressed as $\widehat{f}$ for some $f\in\mathcal{S}(\mathbb{R}^{n})$. The proof of the proposition is now complete since $F^{-1}(0)=E$. ∎

3. Proofs of Theorems 1.1 and 1.2

Let $E$ be a compact set in $\mathbb{R}^{n}$ with induced measure $d\sigma$. Suppose $E$ has the $p$-restriction property. This is equivalent to the existence of a constant $C$ that depends on $p$ and $n$ and satisfies

[TABLE]

for all $f\in\mathcal{S}(\mathbb{R}^{n})$. Condition (3.1) is equivalent to the dual condition

[TABLE]

for all smooth functions $F$ on $E$, and where $\widecheck{Fd\sigma}$ is the inverse Fourier transform of the measure $Fd\sigma$. Recall that the inverse Fourier transform of a finite Borel measure is $d\mu$

[TABLE]

where $x\in\mathbb{R}^{n}$. Setting $F\equiv 1$ on $E$ we see from (3.2) that $\widecheck{d\sigma}\in L^{p^{\prime}}(\mathbb{R}^{n})$. We record this as:

Lemma 3.1**.**

Let $1<p<2$ and let $E$ be a compact set in $\mathbb{R}^{n}$ with induced measure $d\sigma$. If $E$ is a set of $p$-restriction, then $\widecheck{d\sigma}\in L^{p^{\prime}}(\mathbb{R}^{n})$.

Proposition 3.2**.**

Let $1\leq p<2$ and let $E$ be a compact subset of $\mathbb{R}^{n}$ with induced measure $d\sigma$. If $E$ is a set of $p$-restriction, then $I^{p}(E)\neq L^{p}(E)$.

Proof.

Let $f\in I^{p}(E)$ and let $\phi(x)=\widecheck{d\sigma}(x)$. By Lemma 3.1, $\phi(x)\in L^{p^{\prime}}(\mathbb{R}^{n})$. Let $f\in I^{p}(E)$ and let $(f_{n})$ be a sequence of Schwartz functions that satisfy $\|f_{n}-f\|_{L^{p}}\rightarrow 0$. For $x\in\mathbb{R}^{n}$,

[TABLE]

Because $\widehat{f}\mid_{E}=0$ we obtain

[TABLE]

Since $\mathcal{R}_{E}$ has property $\mathcal{R}(E,p,1),\lim_{n\rightarrow\infty}\|\widehat{f}-\widehat{f_{n}}\|_{L^{1}(E)}=0$. Hence, $f\ast\phi=0$ and $\phi$ is a nonzero element in $\text{Ann}(I^{p}(E))$ by Proposition 2.1. Thus $I^{p}(E)\neq L^{p}(\mathbb{R}^{n})$. ∎

The following corollary to Proposition 3.2, which is crucial for the proofs of Theorems 1.1 and 1.2, gives a useful criterion in terms of $T^{p}[f]$ to determine when $E$ is not a set of $p$-restriction.

Corollary 3.3**.**

Let $1\leq p<2$ and let $E$ be a compact subset of $\mathbb{R}^{n}$. If there exists an $f\in L^{1}(\mathbb{R}^{n})\bigcap L^{p}(\mathbb{R}^{n})$ for which $E\subseteq Z(f)$ and $T^{p}[f]=L^{p}(\mathbb{R}^{n})$, then $E$ is not a set of $p$-restriction.

Proof.

Assume $E$ is a set of $p$-restriction. Then by Proposition 3.2 $I^{p}(E)\neq L^{p}(\mathbb{R}^{n})$. Since $Z(f_{y})=Z(f)$ for all $y\in\mathbb{R}^{n},T^{p}[f]\subseteq I^{p}(E)$ which contradicts our hypothesis $T^{p}[f]=L^{p}(\mathbb{R}^{n})$. ∎

3.1. Proof of Theorem 1.1

Theorem 1.1 follows immediately by combining Corollary 3.3 with [1, Corollary 1].

3.2. Proof of Theorem 1.2

We will prove Theorem 1.2 by proving a more general theorem. We start with a definition. Let $E$ be a closed subset of $\mathbb{R}^{n}$. We shall say that $E$ is $p$-thin if the only distribution $T$ that satisfies $\hbox{supp }T\subseteq E$ and $\widecheck{T}\in L^{p}(\mathbb{R}^{n})$ is $T=0$.

Theorem 3.4**.**

Let $1<p<2$ and let $E$ be a compact subset of $\mathbb{R}^{n}$. If $E$ is $p^{\prime}$-thin, then $E$ is not a set of $p$-restriction.

Proof.

Assume that $E$ is $p^{\prime}$-thin. Let $f\in\mathcal{S}(\mathbb{R}^{n})$ with $Z(f)=E$. The theorem will follow from Corollary 3.3 if we can show $T^{p}[f]=L^{p}(\mathbb{R}^{n})$. Suppose instead that $T^{p}[f]\neq L^{p}(\mathbb{R}^{n})$. By Proposition 2.1 there exists a nonzero $\phi\in L^{p^{\prime}}(\mathbb{R}^{n})$ for which $f\ast\phi=0$, which implies $\text{supp }\hat{\phi}\subseteq Z(f)$ because $\widehat{f\ast\phi}=\hat{f}\hat{\phi}$. Due to our assumption $E$ is $p^{\prime}$-thin, $\phi=0$, a contradiction. Hence $T^{p}[f]=L^{p}(\mathbb{R}^{n})$. ∎

It was shown in [6, Theorem 1] that if a set $E$ satisfies the hypothesis of Theorem 1.2 then $E$ is $p^{\prime}$-thin. Therefore, $E$ is not a set of $p$-restriction and Theorem 1.2 is proved.

4. $p$-spectral synthesis

We start with a definition. Let $E$ be a $k$-dimensional submanifold in $\mathbb{R}^{n}$ with induced Lebesgue measure $d\sigma$. We shall say that $E$ has the $p$-approximate property if for each distribution $T$ with $\text{supp }T\subseteq E$ and $\widecheck{T}\in L^{p}(\mathbb{R}^{n})$, we can find a sequence of measures $T_{j}$ on $E$, absolutely continuous with respect to $d\sigma$, such that $\|\widecheck{T_{j}}-\widecheck{T}\|_{L^{p}}\rightarrow 0$ as $j\rightarrow\infty$. Our results on sets of $p$-spectral synthesis are an immediate consequence of previous work by Guo on sets that have the $p$-approximate property [5, 7]. In fact, it is stated in [5] that the $p$-approximate property is a variation of the spectral synthesis property. Sets with the $p$-restriction property allows us to make this statement more transparent. Specifically, Theorem 4.1 will show that $p$-spectral synthesis follows from the $p^{\prime}$-approximate property for submanifolds with the $p$-restriction property.

4.1. Proof of Theorem 1.3

Theorem 1.3 will follow immediately by combining [5, Theorem 1] and [7, Theorem 2] with Theorem 4.1 below.

Suppose $E$ is a $k$-dimensional submanifold of $\mathbb{R}^{n}$ and let $d\sigma$ be the induced Lebesgue measure on $E$. Also assume that $E$ is a set of $p$-restriction for some $1<p<2$. Let $\Phi$ denote the closed subspace of $L^{p^{\prime}}(\mathbb{R}^{n})$ generated by

[TABLE]

Theorem 4.1**.**

Let $1<p<2$ and let $E$ be a compact, smooth $k$-dimensional submanifold of $\mathbb{R}^{n}$ and assume that $E$ has the $p$-restriction property. Let $d\sigma$ be the induced measure on $E$. If $E$ has the $p^{\prime}$-approximate property, then $E$ is a set of $p$-spectral synthesis.

Proof.

Since $E$ is a set of $p$-restriction, $\widecheck{d\sigma}\in L^{p^{\prime}}(\mathbb{R}^{n})$ by Lemma 3.1. We begin by showing showing $\Phi\subseteq\text{Ann}(I^{p}(E))$. Let $\phi\in\Phi$. We can assume that $\phi=\widecheck{d\mu}$, where $d\mu=Fd\sigma$ for some smooth function $F$ on $E$. Let $f\in I^{p}(E)$, using the argument from Proposition 3.2 we obtain that $f\ast\phi=0$, which implies that $\phi\in\text{Ann}(I^{p}(E))$ by Proposition 2.1.

Now let $\phi\in\text{Ann}(\overline{k(E)}^{p})$. Since $\text{supp}(\widehat{\phi})\subseteq E$ and $E$ has the $p^{\prime}$-approximate property, there exists a sequence of measures $F_{n}d\sigma$, where $F_{n}$ is smooth on $E$, such that $\|\widecheck{F_{n}d\sigma}-\phi\|_{L^{p^{\prime}}}\rightarrow 0$. Thus $\phi\in\Phi$, which implies $\text{Ann}(\overline{k(E)}^{p})\subseteq\text{Ann}(I^{p}(E))$. Clearly, $\text{Ann}(I^{p}(E))\subseteq\text{Ann}(\overline{k(E)}^{p})$. Therefore, $E$ is a set of $p$-spectral synthesis. ∎

4.2. $p$-spectral synthesis and the unit sphere

We mentioned in the Introduction that for $n\geq 3,S^{n-1}$ is not a set of spectral synthesis. Using Theorem 1.3 we will be able to show that there are $p$-values for which $S^{n-1}$ is a set of $p$-spectral synthesis. By [5, Theorem 1], $S^{2}$ has the $p^{\prime}$-approximate property when $p^{\prime}>2$ and $n=3$; and $S^{n-1}$ has the $p^{\prime}$-approximate property for $2<p^{\prime}\leq\frac{2n}{n-3}$ when $n\geq 4$. It follows from the Stein-Tomas theorem that $S^{n-1}$ is a set of $p$-restriction for $1\leq p\leq\frac{2n+2}{n+3}$. Consequently, Theorem 4.1 yields that $S^{2}$ is a set of $p$-spectral synthesis for $1<p\leq\frac{4}{3}$ and for $n\geq 4,S^{n-1}$ is a set of $p$-spectral synthesis for $\frac{2n}{n+3}\leq p\leq\frac{2n+2}{n+3}$. The upper bound in this inequality is probably not sharp, in fact it would not surprise us if it is $\frac{2n}{n+1}$. However, the lower bound is sharp. Indeed, using [6, Lemma 2.3(ii)] a distribution $\phi$ can be constructed with $\phi\in L^{p^{\prime}}(\mathbb{R}^{n})\mbox{for }p^{\prime}>\frac{2n}{n-3},\text{supp}(\widehat{\phi})\subseteq S^{n-1}$ that satisfies for $f\in\mathcal{S}(\mathbb{R}^{n})$,

[TABLE]

Since there exists $f\in\mathcal{S}(\mathbb{R}^{n})$ with $\widehat{f}\mid_{S^{n-1}}=0$ and $\frac{\partial\widehat{f}}{\partial x_{n}}\mid_{S^{n-1}}\neq 0,\phi\notin\text{Ann}(I^{p}(E))$. Thus, for $n\geq 4,S^{n-1}$ is not a set of $p$-spectral synthesis for $1\leq p<\frac{2n}{n+3}$.