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Let $K$ be a compact subset in $\mathbb{R}^n$ with $m(K)=0$, Suppose $supp\hat{u}\subset K$ for some $u\in L^p$,where $2\leq p\leq \frac{2n}{n-1}$,can we get $u\equiv 0$ ?

Motivation: If $K$ is a compact non-degenerate hypersurface,then it's well known that $u(x)\leq C|x|^{-\frac{n-1}{2}}$,hence $u\in L^p$ for any $p>\frac{2n}{n-1}$,but if we restrict $p$ in $[2,\frac{2n}{n-1}]$,then is it possible that $u\equiv 0$ ?

Another related question: Consider a linear partial differential operator $P(D)$ with constant coefficients in $L^p$,it has been proven that $P(D)$ has no eigenvalue when $1\leq p<\frac{2n}{n-1}$,the bound for $p$ is best possible,in fact consider $-\triangle$ act on $L^p$, $p>\frac{2n}{n-1}$,it's known that $\sigma(-\triangle)=[0,\infty)$ for all $p$,we now care about its point spectrum(eigenvalue) and it's easy to check that $u=\hat{d\mu}$(where $d\mu$ is the surface measure on {$|\xi|=s^2$})is the eigenfunction of $-\triangle$ with eigenvalue $s^2$,hence,in this case,$\sigma_{p}(-\triangle)=(0,\infty)$.But for $p<\frac{2n}{n-1}$,$\sigma_{p}(-\triangle)=\emptyset$,then what I am particularly interested in is the case when $p=\frac{2n}{n-1}$.Is its point spectrum also empty ?

Edit For the first question,it suffice to show that $\mathcal{F(S_{K})}$ is dense in such $L^p$,where $S_{K}$={$\varphi\in C_{0}^{\infty}(\mathbb{R}^n)$,$\phi(x)=0$,when $x\in K$ },this is obvious when K consists of only discrete points,but for more complicated K(such as a low-dimension surface),I don't know how to deal with it right now.

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up vote 2 down vote accepted

I discussed this problem with fedja (Overflow pen-name), and he explained to me that the answer is no. As fedja is apparently busy, I post the answer:-) One cannot improve the exponent $p=2$ in any dimension. The reason is that for every $p>2$ there exists a distribution on the line whose support has Lebsegue measure $0$ and whose Fourier transform belongs to $L^p$. Now, for arbitrary dimension, one simply takes a product. In Russian, this is called Inashev-Musatov theorem (1957) but this was an improvement of a series of earlier results; apparently the condition $p>2$ is due to Wiener, Amer. J. Math. 60 (1938).

This results were for Fourier series rather than Fourier transform, but here is the reference for Fourier transform: MR0227693.

So one needs stronger assumptions on support than just zero measure. I don't know what these assumptions could be.

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Thanks,one can get the positive answer when adding the condition $m(K_{\delta})<C\delta$,where $K_{\delta}$ is the set {$x\in \mathbb{R}^n$,dist(x,K)$\leq \delta$},for the case $2<p<\frac{2n}{n-1}$,you can see [this paper][1] [1]:… – user23078 Oct 21 '12 at 12:23

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