1 What is Fourier Restriction Theory?

Main object: f:ℝd→ℂ, f(x)=∑⁡ξ∈ℝbξe2πix⋅ξ, bξ∈ℂ.

Notation. We will write e(x⋅ξ)ie2πix⋅ξ.

x∈ℝd is a spatial variable, and ξ∈ℝd 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.

{log⁡n} 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 ℙ(𝜀n=1)=ℙ(𝜀n=−1)=12

  • 1<p<∞

  • x1,…,xN∈ℂ

Then
(𝔼|∑n=1N𝜀nxn|p)1p∼p(∑n=1N|xn|2)12=∥xn∥2.

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

Proof. Without loss of generality, x1,…,xn∈ℝ. Without loss of generality, ∥x∥2=1.

p=2: want to show 𝔼(|∑⁡n𝜀nxn|2)∼1.

𝔼(∑n𝜀nxn∑m𝜀mxm¯)=∑n,m𝔼(𝜀n𝜀mxnxm¯)=∑n|xn|2+∑n∑m≠nxnxm¯𝔼𝜀n⏟=0𝔼𝜀m.

What about general exponents p?

𝔼(|∑n𝜀nxn|p)=∫0∞ℙ(|∑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 eλ∑⁡n𝜀nxn∈(0,∞).

𝔼(eλ∑n𝜀nxn)=𝔼(∏neλ𝜀nxn)=∏n𝔼eλ𝜀nxn=∏n(12eλxn+12e−λxn).

Fact: 12ez+12e−z≤ez22 (to check, use the Taylor series). So we can get

αℙ(eλ∑n𝜀nxn>α)≤𝔼(eλ∑n𝜀nxn)(Chebyshev’s inequality)≤∏neλ2|xn|2∕2=eλ2∕2

By symmetry,

αℙ(e|λ∑n𝜀nxn|>α)≲eλ2∕2.

Choose α=eλ2:

ℙ(|∑n𝜀nxn|>λ)=ℙ(|∑n𝜀nxn|p>λp)≲e−λ2∕2.

Use in Layer cake:

𝔼(|∑n𝜀nxn|p)≲∫0∞e−α2∕p∕2dα∼p1.

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

𝔼(XX¯)⏟=1≤(𝔼(|X|p))1∕p⏟≲p1(𝔼|X|q)1∕q.

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|p∕2dx.

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((n−mx)dx=N).
∫[0,1]|g(x)|2=∑n,m∫[0,1]e((n2−m2)x)dx=N.

PIC

∫[0,1]|f(x)|pdx≥Np∕2+Np−1
∫[0,1]|f(x)|pdx≥Np∕2+Np−2.

For the first one, Np−1 (organised behaviour) dominates as soon as p>2, and for the second one, Np∕2 dominates for 2≤p≤4 (“square root cancellation behaviour lasts for longer”).