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

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.

 (                )
   ∑      ∑-------    ∑          ---   ∑     2   ∑  ∑     ---
𝔼     𝜀nxn   𝜀mxm   =    𝔼(𝜀n𝜀mxnxm ) =   |xn| +       xn xm◟𝔼◝𝜀◜n◞𝔼𝜀m.
    n      m          n,m                n         n m=n      =0
What about general exponents p?
 ( |      | )        ( |       |    )
   ||∑      ||p    ∫ ∞    ||∑       ||p
𝔼  ||  𝜀nxn||   =     ℙ  ||   𝜀nxn || > α  dα.
    n            0      n
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  eλ∑n 𝜀nxn  = 𝔼  ∏  eλ𝜀nxn  = ∏  𝔼eλ𝜀nxn = ∏   1eλxn + 1e−λxn  .
                   n           n           n  2       2
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,

αℙ(e||λ ∑n𝜀nxn|| > α) ≲ eλ2∕2.
Choose α=eλ2:
  (|       |   )     (|       |p    )
   ||∑      ||          ||∑      ||    p     −λ2∕2
ℙ  ||   𝜀nxn|| > λ = ℙ  ||   𝜀nxn|| > λ   ≲ e    .
     n                  n
Use in Layer cake:
 ( ||       ||p)   ∫ ∞
𝔼  ||∑  𝜀 x ||  ≲     e−α2∕p∕2dα ∼  1.
   | n  n n|     0             p
Lower bound: use Hölder’s inequality. X=n𝜀nxn.
    --        p 1∕p     q1∕q
𝔼◟(X◝◜X)◞ ≤ (𝔼◟-(|X◝|◜))-◞(𝔼|X | )  .
  =1        ≲p1
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].

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

PIC

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