1 What is Fourier Restriction Theory?

Main object: f:d, f(x)=ξbξe2πixξ, bξ.

Notation. We will write e(xξ)ie2πixξ.

xd is a spatial variable, and ξd is the frequency variable.

The frequencies (or Fourier transform) of f is restricted to a set R (where R we will always be finite – so no need to worry about convergence issues).

Goal: Understand the behaviour of f in terms of properties of R.

Example.

Both avoid linear structure.

{logn} is a concave set (getting closer and closer together).

{(n,n2)} lie on a parabola.

Guiding principle: if properties of an object avoid (linear) structure, then we expect some random or average behaviour.

The above examples avoid linear structure using some notion fo curvature. See Bourgain Λ(p) paper: extra behaviour.

Square root cancellation: If we add ±1 randomly N times, then we expect a quantity with size N12.

Theorem 1.1 (Khinchin’s inequality). Assuming that:

  • {𝜀n}n=1N be IID random variables with (𝜀n=1)=(𝜀n=1)=12

  • 1<p<

  • x1,,xN

Then
(𝔼|n=1N𝜀nxn|p)1pp(n=1N|xn|2)12=xn2.

Notation. p means but the constant may depend on p.

Proof. Without loss of generality, x1,,xn. Without loss of generality, x2=1.

p=2: want to show 𝔼(|n𝜀nxn|2)1.

𝔼(n𝜀nxnm𝜀mxm¯)=n,m𝔼(𝜀n𝜀mxnxm¯)=n|xn|2+nmnxnxm¯𝔼𝜀n=0𝔼𝜀m.

What about general exponents p?

𝔼(|n𝜀nxn|p)=0(|n𝜀nxn|p>α)dα.

The equality here is the Layer cake formula, which is true for any p(0,).

Let λ>0. Study the random variable eλn𝜀nxn(0,).

𝔼(eλn𝜀nxn)=𝔼(neλ𝜀nxn)=n𝔼eλ𝜀nxn=n(12eλxn+12eλxn).

Fact: 12ez+12ezez22 (to check, use the Taylor series). So we can get

α(eλn𝜀nxn>α)𝔼(eλn𝜀nxn)(Chebyshev’s inequality)neλ2|xn|22=eλ22

By symmetry,

α(e|λn𝜀nxn|>α)eλ22.

Choose α=eλ2:

(|n𝜀nxn|>λ)=(|n𝜀nxn|p>λp)eλ22.

Use in Layer cake:

𝔼(|n𝜀nxn|p)0eα2p2dαp1.

Lower bound: use Hölder’s inequality. X=n𝜀nxn.

𝔼(XX¯)=1(𝔼(|X|p))1pp1(𝔼|X|q)1q.

1p+1q=1.

Can you find a more intuitive proof? E-mail Dominique Maldague.

Corollary. 𝔼(|n=1N𝜀nfn(x)|pdx)p|n=1N|fn(x)|2|p2dx.

Useful for exercises!

Return to Fourier restriction context.

f(x)=n=1Ne(nx)R={1,,N}g(x)=n=1Ne(n2x)R={12,22,,N2}

Both f,g are 1-periodic. So study them on 𝕋=[0,1]. f(0)=N, |f(x)|N for [0,c1N]. g(0)iN, |g(x)|N for x[0,c1N2].

[0,1]|f(x)|2dx=n,m[0,1]e((nmx)dx=N).
[0,1]|g(x)|2=n,m[0,1]e((n2m2)x)dx=N.

PIC

[0,1]|f(x)|pdxNp2+Np1
[0,1]|f(x)|pdxNp2+Np2.

For the first one, Np1 (organised behaviour) dominates as soon as p>2, and for the second one, Np2 dominates for 2p4 (“square root cancellation behaviour lasts for longer”).