Introduction to Fourier Transform

3 Introduction to Fourier Transform

Correction for lecture 2: Number Theory Lemma.

True statements:

and

# \div (n) ≲_{𝜀} n^{𝜀}

See Terence Tao notes online.

Explaining $# \div (n) ≲_{𝜀} n^{𝜀}$ : $\forall 𝜀 > 0$ , $\exists c_{𝜀} \in (0, \infty)$ such that $# \div (n) \leq C_{𝜀} n^{𝜀}$ for all $n \geq 1$ .

We will be using $≾$ to mean “ $\leq$ but up to sub-polynomial in $n$ ”.

Question:

| |p
∫ ||∑ || p− 2 2
− [0,N2]|| bne(anx)|| dx ≾ (1 +N 2 )∥bn∥2.
n∼N

For example,

a_{n} = \frac{n^{2}}{N^{2}}

Recall: $n \sim N$ means $N \leq n \leq 2 N$ .

${a_{n}} \subseteq [0, 10]$ , $a_{n + 1} - a_{n} \sim \frac{1}{N}$ , $(a_{n + 2} - a_{n + 1}) - (a_{n + 1} - a_{n}) \sim \frac{1}{N^{2}}$ .

Reasonable conjecture?

Yes, reasonable.

Khinchin’s inequality: May select $b_{n} \in {\pm 1}$ so that

∫ ||∑ ||p ∫ ||∑ ||p2
− || bne(anx)|| dx ∼ − || |bn|2|| dx.
[0,N2]| n | [0,N2]| n |

Constant integral: $b_{n} \equiv 1$ . $(1, 1, 1, \dots, 1)$ .

| |p
∫ ||∑ || -1-∫ p p−2 p− 2 p
− [0,N2]|| e(anx)|| dx ≳ N 2 [0,c]N dx ∼ N = N 2 ∥bn∥2.
n

Warning. Enemy scenario: ${a_{n}} = {\sqrt{\frac{n}{N}}}_{n = N}^{2 N}$ (technically ${- \sqrt{\frac{3 N - n}{N}}}$ if we want to satisfy the conditions mentioned above).

$(b_{n}) = (1, 0, \dots, 0, 1, 0, \dots, 0, 1, \dots, 0)$ . Length $N$ vector with $N^{\frac{1}{2}}$ many $1$ s. Have:

| ( ∘ --- )|p | |
∫ || ∑ m2 || ∫ || ∑ ( m ) ||p
− 2 || e N-x || dx = − 2 || e √---x ||dx
[0,N ]|N≤m2≤2N | [0,N ] m2∼N N

We can calculate that the above expression is in fact $\geq N^{\frac{p}{2} - \frac{1}{2}}$ (which breaks the conjecture until $p > 6$ ).

It turns out that this is (roughly speaking) the only problem.

Why do we care?

$b_{n} \equiv 1$ , $p = 4$ . Then

∫ ||∑ ||4
− || e(anx )|| dx ≾ N 2 = |{an}|2.
[0,N2]| n |

\to

Convex sequences have minimal additive energy.

Decoupling doesn’t know how to take advantage of $b_{n} \equiv 1$ .

3.1 Fourier Transform on $ℝ^{n}$

$f : ℝ^{n} \to ℂ$ , $f \in S (ℝ^{n})$ Schwartz function: $∥ x^{α} \partial^{β} f ∥_{\infty} < \infty$ for all $α, β$ .

x

is the spatial variable, and

ξ

is the frequency variable.

Facts:

If $f \in S$ , then $\hat{f} \in S$ .
Plancherel’s Theorem: $\int f (x) g (x) d x = \int \hat{f} (ξ) ĝ (ξ) d ξ$ .
$f (x) = e^{- π x^{2}}$ , $\hat{f} (ξ) = e^{- π ξ^{2}}$
$λ > 1$

On the left: mass of $f$ is smashed by a factor of $λ$ . On the right: the mass of $\hat{f}$ is stretched by a factor of $λ$ , “ $L^{1}$ normalized”. $∥ \hat{f_{λ}} ∥_{1}$ is independent of $λ$ .

There is a general formula for $\hat{f \circ A}$ where $A$ is an affine transformation.
$\hat{c f + g} = c \hat{f} + \hat{g}$ .
Translations are dual to modulations:
$\begin{array}{l} f_{τ} (x) & = f (x - τ) \\ \hat{f_{τ}} (ξ) & = e^{- 2 π i ξ \cdot τ} \hat{f} (ξ) \\ f^{t} a u (x) & = e^{2 π i τ \cdot x} f (x) \\ \hat{f^{τ}} (ξ) & = \hat{f} (ξ - τ) \end{array}$

Basic question about $\hat{f}$ : $L^{p}$ to $L^{q}$ boundedness?

Plancherel’s:

(isometry on

L^{2}

$p = 1$ , $q = \infty$ :

||∫ || ∫
|| e−2πix⋅ξf (x)dx|| ≤ |f(x)|dx = ∥f∥1

(contraction from

L^{1}

L^{\infty}

By interpolation (Marcinkiewicz): $∥ \hat{f} ∥_{q} \leq ∥ f ∥_{p}$ for $1 \leq p \leq 2$ , $\frac{1}{p} + \frac{1}{q} = 1$ (Hausdorff-Young inequality).

Are there any other $(p, q)$ for which

Attempt 1: Let $φ \in C_{c}^{\infty} (ℝ^{n})$ (compactly supported smooth function on $ℝ^{n}$ ), with $supp φ \subseteq B_{1} (0)$ .

Consider $f (x) = \sum_{k = 1}^{N} 𝜀_{k} φ (x - v_{k})$ .

Choose $v_{k}$ so that ${B_{1} (v_{k})}$ are not overlapping. Then

∫ ||∑ ||p ∑ ∫
∥f∥pp = || 𝜀kφ(x− vk)|| = |φ(x− vk)|pdx = N ∥φ∥pp.
| k | k

Also

N ( N )
^ ∑ − 2πiξvk ∑ −2πiξvk
f (ξ) = 𝜀ke ^φ(ξ) = 𝜀ke φ^(ξ).
k=1 k=1

We will use Khinchin’s inequality.

$∥ \hat{f} ∥_{q} \sim N^{\frac{1}{2}} ∥ \hat{φ} ∥_{q}$ .

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3 Introduction to Fourier Transform

3.1 Fourier Transform on ℝn

3.1 Fourier Transform on $ℝ^{n}$