Zero Gaussian curvature and restriction estimates 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.
 A: 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. 
