4 Introduction to Fourier Restriction

Theorem (Hausdorff-Young inequality). Assuming that:

  • 1p2

  • 1p+1q=1

Then
f^qfp.

Proof. The inequality is true for p=1, q= and for p=2, q=2 the inequality is true since we have equality (Plancherel).

For values in between we can interpolate.

Are there any other (p,q) for which

f^qfp?(∗)

We saw that 1p2 was necessary (translations / modulations, Khinchin’s inequality).

Scaling: Plug in fλ(x)=f(λx) which is L-normalised (fλ=f). Then fλ^(ξ)=λnf^(λ1ξ) which is L1-normalised (fλ^1=f^1).

(LHS of ())q=n|fλ^(ξ)|qdξ=λnq+n|f^(ξ)|qdξ.
(RHS of ())p=n|fλ(x)|pdx=λn|f(x)|pdx.

So we need for all λ>0:

λn+nqf^qλnpfp.

So we need n+nq=np, i.e. 1p+1q=1.

Classical questions

What is Cp,q the smallest constant such that f^qCp,qfp?

Which functions f satisfy f^qfp=Cp,q?

2014:

f^qCp,qfpdistp(f,maximisers (Gaussian)).

Fourier restriction asks which (p,q) permit estimates f^Lq(R)fLp(n) (R is the restricted frequency set, Rn).

Example. R=B1(0)n, unit ball : Basically, sitll governed by Hausdorff-Young inequality.

Example. R=B1(0){xn=0} (measure 0 subset of n).

RR(pq) means

f^Lq(R)fLp(n)f:n Schwartz

Always: RR(1) true for all R.

In the second example, only this trivial statement is true (i.e. RR(pq) is false for all other values for p,q).

Let R=Sn1 be the unit sphere in n. Consider

f^Lq(Sn1)fLp(n)

where Lq(Sn1) uses the usual surface measure dσ on Sn1.

Notation.

^^ˇx↦→ −xf(x )       ^f(x)       f(x)        f(x)
     ◟◝◜◞        ◟◝◜◞       ◟◝◜◞        ◟◝◜◞
    ∈𝒮(ℝn)      ∈𝒮(ℝn)      ∈𝒮(ℝn)      ∈𝒮(ℝn)

We may call ˇ the “inverse Fourier transform”.

Let φCc(n), -valued, 12 on B1(0), with support in B2(0).

May also assume ˇφ is -valued, ˇφ1 on Bc(0). ˇφ bounded, |ˇφ(x)|m(|x|2+1)mm. |ˇφ(x)| behaves like χB1(0) in Lp.

PIC

Consider dilates fR(x)=ˇφ(R1x)e2πixvr, R1.

Frequency side |fR^(ξ)|RnχBR1(0)(x) (L1-norm).

PIC

Sn1|fR^(ξ)|qdσ(ξ)RnqR(n1)σ(BR1(n)Sn1).

BR1(n)Sn1 cap of radius R1. Spacial side |fR(x)|χBR(0)(x) (in L-norm).

PIC

height 1, n|fR(x)|pdxRn.

Rnn1qRnp, nn1qnp.

Consider: gR(x)=eixwrˇφ(R1x1,,R1xn1,R2xn).

Frequency: |gr^(ξ)||cylinder|1χcylinder(ξ). |cylinder|1| (R1R1R2)1=Rn+1.

PIC

Sn1|gR^(ξ)|qdσ(ξ)R(n+1)qσ(cap of radius R1)=R(n+1)q(n1).

Spatial side: |gR(x)|χcylinder(x) (L-norm).

PIC

|gR(x)|p=Rn1+2=Rn+1.

Rn+1n1qRn+1q, n+1n1qn+1p. Implies B.

On Monday, we will build examples h such that h^ sees all of Sn1.

R=Sn1Rn.

Consider the statement

f^Lq(Sn1)fLp(n).

(recall that we called this RSn1(pq)).

Fix φCc(n), φχB1(0). For computing Lp norms, |φ^|χBc(0).

Wave packet function with localised spatial and frequency behaviour.

Last time: fR(x)=eixvRˇφ(R1x1,,R1xn1,R2xn).

Frequency: |fR^(ξ)|

PIC

Sn1|fR^|qdσ

n+1n1qn+1q

Spatial: |fR(x)|

PIC

1|fR|p

Note: sphere near n looks like (ξ1,112|ξ|2), Sn1suppfR^cap of radius R1.

PIC

Naive attempt: gR(x)=Rnˇφ(Rx)

Frequency: |gR^(ξ)|

PIC

Sn1|gR^(ξ)|qdξ1

Spatial: |gR(x)|

PIC

|gR|pRnRnp.

Deduce: 1Rnp+n, so

p1

(trivial).

|gR^|1 on Sn1 made gR^Lq(Sn1) easy to compute.

Could we improve things?

Could think about R1: then gR^Lq(Sn1)1, gRp1. This is more efficient, but we can’t take a limit. So not so useful.

Build a function H(x) which satisfies |H^(ξ)|1 on Sn1.

Let {𝜃} be a maximal collection of R1-spaced points on Sn1 (#{𝜃}Rn1).

PIC

For each 𝜃, let A𝜃1:nn be an affine map which sends

B1(0)R1××R1×R2 ellipsoid centered at 𝜃, tangent to Sn1 at 𝜃.

Define φ𝜃=φA𝜃, H(x)=𝜃ˇφ𝜃(x).

H^(ξ)=𝜃φ𝜃(ξ)1 on Sn1 (actually on R2-neighbourhood of Sn1).

Sn1|H^(ξ)|qdσ1.

H(x)=𝜃ˇφ𝜃(x).

|φ𝜃(x)| Frequency:

PIC

|ˇφ𝜃(x)| Spatial:

PIC

esssuppH𝜃esssuppˇφ𝜃.

PIC

“bush of tubes”

Rn1 many R××R×R2 tubes in R1-separated directions

n|H(x)|pdx=n|𝜃ˇφ𝜃(x)-valued function|pdx

PIC

Compute

n|𝜃χT𝜃(x)2|p2=n|𝜃χT𝜃(x)|p2=BR|χT𝜃|p2(Rn1)p2Rn+RR2BR|χT𝜃|p2()

Consider overlap of T𝜃 on λSn1 (λ(R,R2)).

Average overlap on λSn1:

λSn1𝜃χT𝜃λ(n1)𝜃λSn1χT𝜃λ(n1)𝜃Rn1=λ(n1)R2(n1).

PIC

Not too hard to check that the number of active T𝜃 on λSn1 is λ(n1)R2(n1).

Now calculate:

()R<λ<R2λ<|x|<2λ|𝜃χT𝜃|p2[λ(n1)R2(n1)]p2λndx.
1R(n+1)p[R2n+R(n1)p2Rn].

Two cases: either R2n dominates or the other term dominates. So p2nn+1.