MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Hi all. My question on M.SE is unanswered ( so I want to post it here. I changed it slightly.

If I have a function $f:\Omega \to \mathbb{R}$ in $H^k(\Omega)$ where $\Omega \subset \mathbb{R}^n$ is a compact surface, then what is known about the Fourier transform $\hat{f}$? What space does it lie in?

Most importantly, I want to know if $\lVert f\rVert_{H^k(\Omega)}$ is equivalent to $\lVert \hat{f}(1+|\xi|^2)^{\frac k 2}\rVert_{L^2(K)}$ for some compact $K$ (recall that this is true if the domains are $\mathbb{R}^n$.)

Does this hold? Thanks for any help.

share|cite|improve this question
It will help if you explain what is $H^k$, and what do you mean by compact surface (how smooth, closed?) – Alexandre Eremenko Dec 10 '12 at 19:00
@AlexandreEremenko $H^k$ is the Hilbert Sobolev space, and by compact surface I mean smooth, and closed without boundary. – maximumtag Dec 13 '12 at 19:54
up vote 2 down vote accepted

This seems to be one special case of the so called Fourier restriction problem (see, for instance, Terry Tao's discussion: If $\Omega$ is a smooth, $(n-1)$-dimensional embedded submanifold (compact or not) of $\mathbb{R}^n$ ($n\geq 2$) without boundary, with induced measure $d\mu$, the measure $f d\mu$ supported in $\Omega$ has its Fourier transform lying in $L^q(\mathbb{R}^n)$ with the bound

$\|(f d\mu)^{\wedge}\|_{L^q(\mathbb{R}^n)}\leq C\|f\|_{L^2(\Omega)}$

only if $q\geq 2\frac{n+1}{n-1}$. As a consequence, no such bound can hold for $f$ as above if $q=2$. This is essentially shown in Lemma 3, page 707 of R. S. Strichartz, "Restrictions of Fourier Transforms to Quadratic Surfaces and Decay of Solutions of Wave Equations", Duke Math. J. 44 (1977) 705-714.

I'm not sure if replacing $L^q(\mathbb{R}^n)$ by $L^q(K)$ for some compact region $K\subset\mathbb{R}^n$ with nonvoid interior allows one to circumvent this "no-go" result, but I rather doubt it. In fact, the $L^p-L^q$ Bernstein inequalities (A.5), page 333 of Terry Tao's book "Nonlinear Dispersive Equations" (American Mathematical Society, 2006), which tell us what happens when we perform a restriction to compact regions in frequency space, go in the opposite direction. That is, you still (apparently) need $q$ to respect the above upper bound even after restricting to $K$.

share|cite|improve this answer
Thanks for the reply. – maximumtag Dec 13 '12 at 19:52

The answer is no, at least for the first two interpretations of your question that I see: I assume that $\Omega$ is a $m$-dimensional compact submanifold of $\mathbb R^n$ with induced Riemannian metric $g$ and volume form $\operatorname\{vol\}(g)$.

  1. Interpretation: $f$ is a distribution of compact support in $\mathbb R^n$ acting on test function $\phi$ by $\langle f,\phi\rangle = \int\_\{\Omega\} \phi.f \operatorname\{vol\}(g)$. Then Fourier transform of $f$ is a tempered distribution, not rapidly decreasing since $f$ has singular support.

  2. Interpretation: $f$ is a cocurrent. It acts on smooth $m$-forms $\omega$ with compact support as $\langle f,\omega\rangle = \int\_\{\Omega\} f.\omega$. Here your Sobolev index has to be $k> \dim(\Omega)/2 +1$ so that $f.\omega$ is at least a $C^1$ form.

  3. Interpretation: If $\Omega$ is a symmetric space of certain type isometrically embedded in $\mathbb R^n$ there might be a (spherical) Fourier transform for functions on $\Omega$ itself. This is quite subtle.

share|cite|improve this answer
Thanks for replying. I am not used to these terms that you have in your interpretation.. $f$ is just a function defined on a hypersurface. – maximumtag Dec 13 '12 at 19:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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