Natural of the restriction problem



the most natural problem in harmonic analysis may be:

investigate for what pair $latex (p,q)$ we have :

$latex L^p(R^n)\longrightarrow L^q(R^n)$

$latex \hat f(x)=\int_{R^n}e^{-2\pi ix\xi}f(\xi)d\xi$

is strong-$latex (p,q)$ bounded.

obvious we have the paserval identity:$latex ||\hat f||_{2}=||f||_2$,and we have $latex ||\hat f||_{\infty}\leq||f||_{1}$.

so by the Riesz-Thorin inteplotation theorem we have the Hausdorff-Young inequality:

$latex \forall 1\leq p\leq 2,\frac{1}{p}+\frac{1}{q}=1$ we have:

$latex ||\hat f||_{q}\leq ||f||_{p}$.

now let talk about the rescaling trick:

consider the transform:$latex f(x)\longrightarrow f(\frac{x}{\lambda})=f_{\lambda}(x)$.we know if the inequality is right then it is necessary to have the same growth for the RHS and LHS.

this argument will derive:$latex \frac{n}{p}=n-\frac{n}{q}$.

in fact $latex ||f_{\lambda}(x)||_{p}=\lambda^{\frac{n}{p}}||f(x)||_{p}$.

by the variable substitute formula:$latex \hat f_{\lambda}(x)=\int_{R^n}e^{-2\pi ix\xi}f(\frac{\xi}{\lambda})d\xi=\lambda^n\hat f(\lambda x)$.

so $latex ||\hat f_{\lambda}(x)||_{q}=\lambda^{n-\frac{n}{q}}||\hat f(x)||_{q}$

and by the scaling invariance trick we know the pair $latex (p,q)$ should live on the line $latex \frac{1}{p}+\frac{1}{q}=1$,and by test with the guessian function $latex g(x)=e^{-x^2}$ we know the right pair  should be $latex 1\leq p\leq 2$.this end the problem with $latex R^n$.


now replace $latex R^n$ by a bounded open set $latex K$.when the fourior transform restriction on $latex K$ is bounded $latex p-q$ operator?

i.e. $latex L^p(R^n)\longrightarrow L^q(R^n)$

$latex f\longrightarrow \hat f|_{K}$.

$latex \hat f|_{K}=\int \chi_{K}e^{2\pi i<x,\xi>}f(\xi)d\xi$.

on a bounded set $latex K$,we always have:if $latex q\geq r$,$latex ||f||_{L^p(K)}\geq ||f||_{L^r(K)}$.

and associate with  hausdorff-young inequality we have:

$latex ||\hat f|_K||_{r}\leq ||\hat f||_q\leq ||f||_p$.

and this area is the exact area(rescaling trick and test with gaussian function),so end of the story.(but why?)


Now we begin to deal with the really interesting case:$latex K$ is not a open set but a sub manifold like the unit sphere $latex S^{n-1}$.

$latex ||\hat f||_{L^q(S^{n-1})}\leq ||f||_{L^p(R^{n})}$.

$latex S^{n-1}$ equip with the usual surface measure $latex \sigma$.

but the inequality is not always meaningful.

case:$latex p=2,\hat f\in L^2$,in general can not restrict to a measure zero set due to the loss of regularity.

case:$latex p=1$,$latex \hat f$ continuous,meaningful to restrict to $latex S^{n-1}$.

$latex ||\hat f||_{L^{\infty}(S^{n-1})}\leq ||f||_{L^1(R^n)},\forall 1\leq p\leq \infty$.

Duality:we use the duality argument to transform the “restriction theorem” to “extension theorem”.

$latex T:f\longrightarrow \hat f$.

$latex T:f\longrightarrow \hat f$.

$latex ||f||=\sup_{||f||_p=1}||\hat f||_q=\sup_{||f||_p=1}\sup_{||g||_{q’}=1}|\int_{R^n}\hat fgd\sigma|=\sup_{||g||_{q’}=1}\sup_{||f||_p=1}|\int_{R^n}\hat fgd\sigma|=\sup_{||g||_{q’}=1}||\hat{gd\sigma}||_{p’}$.


we use $latex R_s(p\to q)$ to state the estimate $latex ||\hat f||_{L^q(S)}\leq ||f||_{L^p(R^n)}$.$latex S=S^{n-1}$.

and by rescaling argument we have natural condition:$latex p<\frac{2n}{n-1}$,$latex p’\geq \frac{n+1}{(n-1)q}$.the restriction conjecture just say this necessary condition is also enough.

Now we state the Tomas-Stein restriction theorem:

$latex 1\leq p\leq \frac{2n+1}{n+3}$.$latex R_s(p\to 2)$ holds.

this is the endpoint estimate in dimension 2 case,so by Meceztaze interpolation theorem this lead to the whole restriction theorem in dimension 2.

the first argument is come from the so called $latex TT^*$ trick that is find by fefferman and stein in 1970.

$latex T$ bdd $latex p\to 2$ $latex \Longleftrightarrow$ $latex TT^*$ bdd $latex p’\to p$.

in fact:

$latex ||T||=\sup_{||f||_p=1}||Tf||_2=\sup_{||f||_p=1}\sup_{||g||_2=1}|\int (Tf)g|=\sup_{||f||_p=1}\sup_{||g||_2=1}|\int f(Tg)|=\sup_{||g||_2=1}||Tg||_{p’}=||T^*||$.

$latex \int|e^{2\pi ix\xi}f(\xi)|^2 dw(\xi)\leq C||f||_p^2$

$latex <\hat f,\hat fw(\xi)>\leq c||f||_p^2$

$latex <\hat f,\hat{f*\hat{w(\xi)}}> \leq ||f||_p^2$

$latex <f,f*\hat{w(\xi)}>\leq ||f||_p^2$

$latex <f,f*\hat{w(\xi)}>\leq ||f||_p||f*\hat{w(\xi)}||_{p’}$

this can be derived from HLS inequality:

$latex ||f*\hat{w(\xi)}||_{p’}\leq ||f||_p$.