Restriction Theorem 1

02-09

the most natural problem in harmonic analysis may be:

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

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

$hat f(x)=int_{R^n}e^{-2pi ixxi}f(xi)dxi$

is strong $p-q$ bounded.

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

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

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

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

now let talk about the rescaling trick:

consider the transform: $f(x)longrightarrow f(frac{x}{lambda})=f_{lambda}(x)$ .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: $frac{n}{p}=n-frac{n}{q}$ .

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

by the variable substitute formula: $hat f_{lambda}(x)=int_{R^n}e^{-2pi ixxi}f(frac{xi}{lambda})dxi=lambda^nhat f(lambda x)$ 。

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

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

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

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

$flongrightarrow hat f|_{K}$ .

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

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

and associate with hausdorff-young inequality we have:

$||hat f|_K||_{r}leq ||hat f||_qleq ||f||_p$

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: $K$ is not a open set but a sub manifold like the unit sphere $S^{n-1}$ .

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

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

but the inequality is not always meaningful.

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

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

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

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

$T:flongrightarrow hat f$ .

$||f||=sup_{||f||_p=1}||hat f||_q=sup_{||f||_p=1}sup_{||g||_{q}=1}|int_{R^n}hat fgdsigma|=sup_{||g||_{q}=1}sup_{||f||_p=1}|int_{R^n}hat fgdsigma|=sup_{||g||_{q}=1}||hat{gdsigma}||_{p}$ .

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

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

Now we state the Tomas-Stein restriction theorem:

$1leq pleq frac{2n+1}{n+3}. R_s(p o 2)$ 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 $TT^*$ trick that is find by fefferman and stein in 1970.

$T$ bdd $p o 2$ $Longleftrightarrow TT^* bdd p o p$ .

in fact:

$||T||=sup_{||f||_p=1}||Tf||_2=sup_{||f||_p=1}sup_{||g||_2=1}|int (Tf)g|=sup_{||f||_p=1}sup_{||g||_2=1}|int f(Tg)|=sup_{||g||_2=1}||Tg||_{p}=||T^*||$ .