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.