3 Introduction to Fourier Transform

Correction for lecture 2: Number Theory Lemma.

True statements:

and $\#÷(n){\lesssim }_{𝜀}{n}^{𝜀}$.

See Terence Tao notes online.

Explaining $\#÷(n){\lesssim }_{𝜀}{n}^{𝜀}$: $\forall 𝜀>0$, $\exists c_{𝜀}\in (0,\infty )$ such that $\#÷(n)\le C_{𝜀}{n}^{𝜀}$ for all $n\ge 1$.

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

Question:

$$-{\int }_{[0,N^2]}{\left|\sum _{n\sim N}{b}_{n}e(a_{n}x)\right|}^{p}dx\precsim (1+N^{\frac{p}{2}-2})\parallel b_n{\parallel }_2^2.$$

For example, $a_n=\frac{n^2}{N^2}$.

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

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

Reasonable conjecture?

Yes, reasonable.

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

$$-{\int }_{[0,N^2]}{\left|\sum _n{b}_{n}e(a_{n}x)\right|}^{p}dx\sim -{\int }_{[0,N^2]}{\left|\sum _n|b_n{|}^2\right|}^{\frac{p}{2}}dx.$$

Constant integral: $b_n\equiv 1$. $(1,1,1,\dots ,1)$.

$$-{\int }_{[0,N^2]}{\left|\sum _{n}e(a_{n}x)\right|}^{p}dx\gtrsim \frac{1}{N^2}{\int }_{[0,c]}{N}^{p}dx\sim N^{p-2}=N^{\frac{p}{2}-2}\parallel b_n{\parallel }_2^p.$$

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

Why do we care?

$b_n\equiv 1$, $p=4$. Then

$$-{\int }_{[0,N^2]}{\left|\sum _{n}e(a_{n}x)\right|}^{4}dx\precsim N^2=|\{a_n\}{|}^2.$$

$$⟹\#\{a_{n_1}+a_{n_2}=a_{n_3}+a_{n_4}\}\lesssim |\{a_n\}{|}^2.$$

$\to$ 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: $\parallel x^{\alpha }{\partial }^{\beta }f{\parallel }_{\infty }<\infty$ for all $\alpha ,\beta$.

$$\widehat{f}(\xi )={\int }_{{ℝ}^n}{e}^{-2\pi ix\cdot \xi }f(x)dx.$$

$x$ is the spatial variable, and $\xi$ is the frequency variable.

Facts:

• If $f\in S$, then $\widehat{f}\in S$.

• Plancherel’s Theorem: $\int f(x)\overline{g(x)}dx=\int \widehat{f}(\xi )\overline{ĝ(\xi )}d\xi$.

• $f(x)=e^{-\pi x^2}$, $\widehat{f}(\xi )=e^{-\pi {\xi }^2}$

• $\lambda >1$

  

  On the left: mass of $f$ is smashed by a factor of $\lambda$. On the right: the mass of $\widehat{f}$ is stretched by a factor of $\lambda$, “$L^1$ normalized”. $\parallel \widehat{f_{\lambda }}{\parallel }_1$ is independent of $\lambda$.

  There is a general formula for $\widehat{f\circ A}$ where $A$ is an affine transformation.

• $\widehat{cf+g}=c\widehat{f}+\widehat{g}$.

• Translations are dual to modulations:

  $$\begin{array}{llll}\hfill f_{\tau }(x)& =f(x-\tau )& \hfill & \\ \hfill \widehat{f_{\tau }}(\xi )& =e^{-2\pi i\xi \cdot \tau }\widehat{f}(\xi )& \hfill & \\ \hfill f^{t}au(x)& =e^{2\pi i\tau \cdot x}f(x)& \hfill & \\ \hfill \widehat{f^{\tau }}(\xi )& =\widehat{f}(\xi -\tau )& \hfill & \end{array}$$

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

Plancherel’s:

$$\parallel \widehat{f}{\parallel }_2^2=\int \widehat{f}\overline{\widehat{f}}=\int f\overline{f}=\parallel f{\parallel }_2^2$$

(isometry on $L^2$).

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

$$\left|\int e^{-2\pi ix\cdot \xi }f(x)dx\right|\le \int |f(x)|dx=\parallel f{\parallel }_1$$

(contraction from $L^1$ to $L^{\infty }$).

By interpolation (Marcinkiewicz): $\parallel \widehat{f}{\parallel }_q\le \parallel f{\parallel }_p$ for $1\le p\le 2$, $\frac{1}{p}+\frac{1}{q}=1$ (Hausdorff-Young inequality).

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

$$\parallel \widehat{f}{\parallel }_q{\lesssim }_{p,q}\parallel f{\parallel }_q?$$

Attempt 1: Let $\phi \in C_c^{\infty }({ℝ}^n)$ (compactly supported smooth function on ${ℝ}^n$), with $\mathrm{supp}\phi \subseteq B_1(0)$.

Consider $f(x)={\sum }_{k=1}^N{𝜀}_k\phi (x-v_k)$.

Choose $v_k$ so that $\{B_1(v_k)\}$ are not overlapping. Then

$$\parallel f{\parallel }_p^p=\int {\left|\sum _k{𝜀}_k\phi (x-v_k)\right|}^p=\sum _k\int |\phi (x-v_k){|}^{p}dx=N\parallel \phi {\parallel }_p^p.$$

Also

$$\widehat{f}(\xi )=\sum _{k=1}^N{𝜀}_k{e}^{-2\pi i\xi v_k}\widehat{\phi }(\xi )=\left(\sum _{k=1}^N{𝜀}_k{e}^{-2\pi i\xi v_k}\right)\widehat{\phi }(\xi ).$$

We will use Khinchin’s inequality.

$\parallel \widehat{f}{\parallel }_q\sim N^{\frac{1}{2}}\parallel \widehat{\phi }{\parallel }_q$.