6 Tube incidence implications of Fourier restriction

Last time: LOCALLY CONSTANT PROPERTY (think of it more as “heuristic”).

If suppf^B1 then

What if suppf^Bλ(0) (so |f|const on λ1-balls)?

f=ˇ(f^φBλ)=fˇφBλ,

where φBλ(ξ)=φB1(λ1ξ).

ˇφBλ(x)=ne2πixξφB1(λ1ξ)dξ=λnne2πi(λx)ξφB1(ξ)dξ=λnˇφB1(λx).

|ˇφB1| is approximately averaging over a 1-ball. |ˇφBλ| is approximately averaging over λ1-balls.

What about suppf^Bλ(v¯)?

Same thing happens, because eixsomethingf will have Fourier support in Bλ(0), and taking absolute values means we don’t notice the eixsomething (modulation).

Returning to RPn1(qp)RPn1,loc(qp).

BR|f(x)|pdxBR|f(x)φBR(x)|pdx

φBR(x)S(n), |φBR|1 on BR, suppφBR^BR1(0).

Last lecture

IR1(|ξ|<1ξn1|f^φBR^(ξ,|ξ|2+ξn)|qdξ)()(R1)1pq(ξn|f^(ξ)|qdξ)pq.

Can choose φBR^ such that

Case 1: pq1. LHS of () (using Hölder) is

|IR1|1pq(IR1|ξ|<1(|f^||φBR^|1q+1q(ξ,|ξ|2+ξn))qdξdξn)pq|IR1|1pq(n(|f^||φBR^|(ξ))qddξ)pq(|f^|q(ξη)|φBR^|(η)dη)1q(|φBR^|qq)1q1(Holder)|IR1|1pq(nn|f^|q(ξη)|φBR^|(η)dηdξ)pq|IR1|1pq(n|f^|q(ξ)dξ)pq

Case 2: pq>1. Use P12 for intuition.

PIC

Imagine a function g which is approximately constant on each R1 cube Q. Think of g as g=QgQ.

IR1(|t|<1g(t,t2+ξ2)dt)pqdξ2.

Note

|t|<1g(t,t2+ξ2)dtgQ.

Therefore,

|t|<1g(t,t2+ξ2)dtC

if |ξ2|cR1. (C for all ξ2IR1).

|(P+(0,ξ2))Q|R1 for |ξ2|cR1.

|IR1|1pq+pqCpq=|IR1|1pq(|IR1|C)pq(R1|t|<1g(t,t2+ξ2)dtdξ2)pq

Important: locally constant property means we didn’t need pq<1, like before.

Make the intuition rigorous.

LHS of ()|IR1|maxξ2IR1(|ξ|<1(|f^||φBR^|(ξ,|ξ|2+ξn))qdξ)pq

Consider the integral:

|ξ|<1(|f^||φBR^|(ξ,|ξ|2+ξn))qdξ=|ξ|<1(n|f^|(η)|φBR^|((ξ,|ξ|2+ξn)η)dη)qdξ|ξ|<1(n|f^|q(η)|φBR^|((ξ,|ξ|2+ξn)η)dη)dξSame pointwise HolderRn(R1)n1n|f^|q(η)dηR1n|f^|q(R1)1Rpq(|f^|q)pq

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