The following is the proof of Lemma 10.25 (page 388) from Tao and Vu's Additive combinatorics, which I'm stuck at and I could really use some help.
The proof begins by taking $P=[1,N]$ and then we use kronecker's approximation to find an integer $1\le r \le N^{1/2}$ such that $||r\xi||_{\mathbb{R} / \mathbb{Z}} \le N^{-1/2}$. Now let $P_0$ denote the progression $[1, \sigma N^{1/2}/100].r$. Then the proof goes onto show that (this equality below is not clear to me) $$|\sum_{n} \mathbb{E}_{x\in P_0} f(n+x)e(n\xi)e(x\xi)|=|\sum_{N}f(n)e(n\xi)| \ge\sigma N$$
Can someone please help me in understanding this equality?
Also, in the very next step, we have that since $x \in P_0$, we have $|e(\xi x)-1|\le \sigma/10$, and so $$|\sum_n \mathbb{E}_{x\in P_0}f(n+x)e(n\xi)(e(x\xi)-1)|\le \sum_{n\in [-N,N]}\sigma/10$$ How does this inequality step follow, it seems that the expression survives only on $[-N,N]$?
I have been trying to work through this, but I am unable to understand these steps.
thanks
The sum goes over all $n \in \mathbf{Z}$, so $$ \sum_n f(n+x) e(n \xi) e(x \xi) = \sum_n f(n) e(n \xi)$$ for all $x \in \mathbf{Z}$. Hence by averaging over $x \in P_0$ we get $$ \sum_n \mathbb{E}_{x \in P_0} f(n+x) e(n\xi) e(x\xi) = \sum_n f(n) e(n\xi).$$ Then take absolute values.
Indeed, $\mathbb{E}_{x \in P_0} f(n+x) \times (\text{other stuff})$ can be nonzero only if there is some $x \in P_0$ such that $n + x \in P$, or in other words only if $n \in P - P_0$. Note that $P - P_0 \subset [-N, N]$.
Does that help? Do also bear in mind the large number of errata listed on Tao's website: https://terrytao.wordpress.com/books/additive-combinatorics/