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Let $$ \left(\int_M|\hat{f}|^qd\mu(\xi)\right)^{1/q}\leq c||f||_{L^p(\mathbb{R}^n)} $$ be a restriction estimate for a hypersurface $M\subset\mathbb{R}^n,~1<p<\infty$ and $\mu$ a surface measure. If we want nontrivial results when analysing this inequality, then the hypersurface $M$ must have some Gaussian curvature. Why is this? What happens if $M$ has zero Gaussian curvature? Specifically, what is the connection between the Gaussian curvature of the hypersurface, and establishing the restriction estimate?

The above inequality can be read as "we can apply the surface Fourier transform to the hypersurface under $L^q$ norm and extend it to $f$ on $\mathbb{R}^n$". So why does the Gaussian curvature of $M$ matter?

Thank you in advance for your help.

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    $\begingroup$ I am going to assume that all we know is that $f\in L^p(\Bbb R^n)$. Perhaps because a zero Gaussian curvature Hypersurface is locally isometric to a plane, and thus compared to $\Bbb R^n$ is a set of measure zero. Consider the case $p=q=2$, then $\|\hat f\|_{L^2(M)}=\|f\|_{L^2(M)}=0$. So in a sense it is not that the statement is not true, it is just that it would be trivial. $\endgroup$ – Ellya Aug 27 '15 at 20:29
  • $\begingroup$ When we have zero Gaussian curvature, we get trivial results and that would make sense. $\endgroup$ – user230715 Aug 28 '15 at 8:32
  • $\begingroup$ @ellya: there is a reason why the statement includes the requirement that $\mu$ is a surface measure. Surface measures are singular relative to the Lebesgue one, so certainly $\|\hat{f}\|_{L^2(M,\mu)}$ does not necessarily vanish. $\endgroup$ – Willie Wong Sep 2 '15 at 2:54
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For illustration let us consider the case where the ambient space is $\mathbb{R}^3$ and $M$ is the $(x,y)$-plane (which is a surface of 0 Gaussian curvature).

Let $g$ be a function of compact support in $\mathbb{R}^3$ whose support include the origin. Let $g_{\nu}(x,y,z) = g(x, y, z/\nu)$. Its inverse Fourier transform scales like $$ \check{g}_{\nu}(x,y,z) =\nu \check{g}(x, y, \nu z) $$ So $$ \| \check{g}_{\nu} \|_{L^p} = \nu^{1 - \frac{1}{p}} $$

Let $\mu$ be the natural induced surface measure on $M$.

Now we have that $\|g_{\nu}\|_{L^q(M,\mu)} = \|g\|_{L^q(M,\mu)} $ since the restriction of $g|_M$ is not changed by scaling the $z$ direction. This contradicts any estimate of the form $$ \|g_{\nu}\|_{L^q(M,\mu)} \leq c \|\check{g}_{\nu}\|_{L^p} \approx \nu^{1 - 1/p} $$ with $p > 1$ since as $\nu \to 0$ the right hand side can be made arbitrarily small but the left hand side is a fixed constant.

This shows you the worst case scenario of what can go wrong.

The case $p = 1$ is special, since we have the trivial estimate from the definition of the Fourier transform that $\|\hat{f}\|_{L^\infty} \leq \|f\|_{L^1}$ and $L^\infty$ behaves quite well after restriction to surface measure.

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