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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$$E(\|\sum_{k=1}^N\epsilon_kf_k\|_p^p)\sim \|(\sum_{k=1}^N|f_k|^2 )^{1/2}\|_p^p,$$ 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?

  • 0
    What is exactly meant by choice of signs?2012-07-17
  • 0
    just that I didn't understand. I think there are measurable functions.2012-08-05

1 Answers 1