At the end of page 5 of the Tao's lectures notes, he sets $\psi$ a Schwartz function supported on the unit cube $[0,1]^n$ and choose $f(x)=\sum_{k=1}^N\epsilon_k\psi(x-ke_1)$, where $e_1$ is one of the basis vectors of $\mathbb{R}^n$, $N$ is a large integer and $\epsilon_k$ are a collection of independent identically distributed signs $\epsilon_k=\pm1$. We have $\|f\|_p\sim N^{1/p}$ (here $A\sim B\Leftrightarrow A\lesssim B$ and $B\lesssim A$, where $A\lesssim B$ means that there is C such that $A\leq C.B$.) Using the Khinchin inequality: If $f_1,\ldots,f_N$ are a collection of functions and $\epsilon_k$ are randomized signs, then for any $1 we have $E(\|\sum_{k=1}^N\epsilon_kf_k\|_p^p)\sim \|(\sum_{k=1}^N|f_k|^2 )^{1/2}\|_p^p,$ where the constants in the $\sim$ symbol are independent of $N$ and $f_k$ and $E$ denotes the expectation, we see that $E(\|\hat{f}\|_q^q)\sim \|(\sum_{k=1}^N|\hat{\psi}(\xi)e^{2\pi ik\xi_1}|^2)^{1/2}\|_q^q\sim N^{q/2}.$ I don't understand the following statement: there must exist some choice of signs for which $\|\hat{f}\|_q\gtrsim N^{1/2}$. Help me, please?
Using Khinchin's inequality
3
$\begingroup$
probability
analysis
fourier-analysis
-
0just that I didn't understand. I think there are measurable functions. – 2012-08-05
1 Answers
3
The signs $\epsilon_k$ are not functions of $x$; they are constants $1$ or $-1$. You can avoid the entire discussion of them being random, independent, etc. by writing $2^{-N}\sum_{\epsilon\in \{-1,1\}^N}(\cdots)$ everywhere you see $E(\cdots)$. The punchline is that if the average of some quantity $Q(\epsilon)$ over all epsilons is $A$, then there exists a value $\epsilon_0$ ("choice of signs") for which $Q(\epsilon_0)\ge A$. Which is exactly the conclusion here: $\|\hat f\|_q^q\gtrsim N^{q/2}$.