On bounding an integral from below.

Question

I am trying to read Exploring the toolkit of Jean Bourgain, a beautiful article by Terrence Tao, that can be found here. At page 5 Tao considers the integral

$$ \int_{|\xi|\le \delta/ t_j}| \hat{1}_B(\xi) |^2 \hat{\sigma}(t_j\xi) d \xi \tag{1} $$

where $0< \delta= \delta(\epsilon) \le 1/2$ is a small quantity depending on $0<\epsilon< 1/2$ to be chosen later, $B \subset [-1,1]^2$, $0<t_j \le 1$, $\hat{1}_B$ is the Fourier transform of the indicator function $1_B$, i.e.,

$$ \hat{1}_B(\xi) = \int_{\mathbb{R}^d} 1_B(x) e^{2\pi i x \xi} dx , $$

and $\hat{\sigma}(\xi) = \int_{S^1} e^{-2\pi i \omega \xi} d \sigma( \omega) $ is the Fourier transform of the surface measure $d \sigma$ on the unit circle $S^1$.

Tao states that, for $\delta$ small enough the factor $\hat{\sigma}(t_j\xi)$ is close to $1$ and thus it is not difficult to show that the integral in $(1)$ is approximately $\gtrsim|B|$, thus

$$ \int_{|\xi|\le \delta/ t_j}| \hat{1}_B(\xi) |^2 \hat{\sigma}(t_j\xi) d \xi \gtrsim |B| \ge \epsilon^2 . $$

How does one rigorously show this?

Answer 1

I just looked at Tao's paper, and he has the lower bound as $|B|^2$.

Given $\epsilon > 0$, there exists $\delta > 0$ such that if $|\xi| < \delta$, then for $x \in [-1,1]^2$ $$ | e^{2 \pi i x \xi} - 1 | < \epsilon .$$ Then if $|\xi| < \delta/t_j$, then $$ \hat \sigma(t_j \xi) = \text{Re}\int \exp(2 \pi i t_j x \xi) \, d\sigma(x) > (1 - \epsilon) \ge 0 $$ and if $|\xi| < \delta$ then $$ |\hat 1_B(\xi) - |B| | = \left|\int_B \exp(2 \pi i x \xi) - 1 \, dx \right| < \epsilon |B| .$$

Hence $$ \int_{|\xi| < \delta/t_j} |\hat1_B(\xi)|^2 \hat\sigma(t_j\xi) \, d\xi \ge \int_{|\xi| < \delta} |\hat1_B(\xi)|^2 \hat\sigma(t_j\xi) \, d\xi \ge \pi \delta^2 (1-2\epsilon)^2 |B|^2 .$$

A similar argument shows that $|B|^2$ is also a good upper bound.