2 Exponential sums in Lp

Recall: studying f(x)=ξRe(ξx). When R does not have (linear) structure, expect |f(x)||R|12 (“sqrt cancellation”) in an appropriate sense (in Lp), more than when R is structured.

Linear R f(x)=n=1Ne(nx) vs convex R g(x)=n=1Ne(n2x).

Have

∫                  ∫
     |f(x)|2dx = N =      |g(x)|2dx.
 [0,1]               [0,1]
Have |f(x)|N on [0,cN1], and |g(x)|N on [0,cN2].

PIC

L2 does not distinguish f,g. L does not distinguish. However, 2<p< does.

What about 1p2? We don’t usually study this range because the estimates tend to be trivial / not interesting.

Focus on 2<p<. Preduct size of

∫      p
 [0,1]|f|.

Square root cancellation lower bound:

     ∫            ( ∫        ) 2p
N  =     |f |2 H öl≤der’s     |f|2⋅p2   (1)□
      [0,1]            [0,1]
Np2[0,1]|f|p.

Constant integral lower bound:

[0,1]|f|p[0,cN1]|f|pNpN1[0,1]|g|p[0,cN2]|g|pNpN2

|f|pNp2+Np1, |g|pNp2+Np2.

Note that for the f bound, Np1 is bigger than Np1 (as long as p>2), so the constant integral dominates!

For g, if 2p4, the square root cancellation dominates, but for p>4 the constant integral takes over.

Assuming:

Theorem 2.1. - p>2 - bn Then:

∫ ||∑        ||p
  ||   bne(nx)|| ≤ N p2−1∥bn∥p2.
  | n       |

Proof. Consider: n=1Nbne(nx), bn.

[0,1]|nbne(nx)|pdxnbne(nx)p2[0,1]|nbne(nx)|2dx(N12bn2)p2bn22CS=Np21bn2p

Note that this is sharp when bn1.

n=1Nbne(n2x). Focus on p=4.

    |          |4
∫   ||∑       2 ||      ∑  ∑  ∑  ∑       ------∫       2   2   2    2
 [0,1]||   bne(n x)|| dx =             bn1bn2bn3bn4 [0,1]e((n1 + n2 − n3 − n4)x)dx.
      n                n1  n2 n3 n4
The integral vanishes unless n12n32=n42n22.

Number Theory lemma: If m, then

# divisors of m ≲ logm.
Follows from unique prime factorisation.

Warning. The above lemma is false.

See correction later.

For fixed n1,n3,

            2   2
# {◟(n2,n4) : n1 −-n3-=◝◜(n4 +-n2)(n4-−-n2)}◞≲ log N.
                =:𝒮n1,n3
Hence

[0,1]|nbne(n2x)|4dxn1n3|bn1bn3||(n2,n4)Sn1,n3bn2bn4|(logN)bn24

We will now use to mean up to powers of logN.

2<p<4: 1p=𝜃2+1𝜃4, 𝜃[0,1], h=nbne(n2x), hph2𝜃h41𝜃bn2𝜃bn21𝜃bn2.

p4:

∫
      p      p− 4   4 CS   12     p−4    4    p2−2    p
 [0,1]|h| ≾ ∥h∥∞ ∥bn∥2 ≤ (N  ∥bn∥2)  ∥bn∥2 ≈ N   ∥bn∥2.

Assuming:

Theorem 2.2. - 2>p Then:

∫ ||∑          ||p
  ||   bne(n2x)|| dx ≾ (1 = N p2− 2)∥bn∥p2.
  |n         |

Sharp by bn1.

Positive take away: estimates are sharp, proofs are elementary. Easy to think of sharp examples.

Number Theory counting idea shows:

[0,1]2|u(x,t)|6dxdtbn26u(x,t)=|n|Nnbne(nx+n2t)[0,1]3|u(x,t)|4dxdtbn24u(x,t)=|n|Nn2bne(nx+|n|2t)

Sharp, 6,4=pcrit.

Strichartz estimate for periodic Schrödinger equation, observed by Bourgain in 1990s.

Negatives: on 𝕋3 pcrit=103, but this technique can only work on even integer values of p.

𝕋1, 𝕋2 only sharp Strichartz estimates per Schrödinger until 2015!

2015: Bourgain-Demeter proved (l2,Lp) sharp decoupling estimate. Gives sharp Strichartz estimate for Schrödinger in 𝕋d for all d.

Proved earlier:

          |           |
    ∫     ||∑    ( n2 )||p           p      p
N −2    2 ||   e  N-2x || dx ≤ (1+ N 2−2)∥bn ∥2.
     [0,N ] n∼N
(where nN means Nn2N).

Conjecture:

  ∫    |           |p
−      ||∑   b e(a x)|| dx ≲ (1 + N p2−2)∥b ∥p
   [0,N2]||n∈N  n  n  ||                 n 2

an[0,1], an+1an1N, an+2an+1(an+1an)1N2.

Example: {an}{n3N3}n=N2N.