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.


1 Answer 1


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.

  • $\begingroup$ 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]: projecteuclid.org/… $\endgroup$
    – user23078
    Oct 21, 2012 at 12:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.