The following question (statement) is from Tao and Vu's Additive combinatorics, at which I am stuck and would appreciate some help.
I wish to prove Theorem 10.27 (page 390) using Proposition 10.28 and Lemma 10.29.
Proposition 10.28: Let $A \subset [1,N]$ be such that $|A| = \delta N$ for some $0<\delta\le 1$ and such that $A$ has no proper arithmetic progressions of length $3$. Suppose also that $N \ge 100/(\delta)^2$. Let $p$ be a prime between $N$ and $2N$, and identify $[1,N]$ with a subset of $\mathbb{Z}_p$. Let $f_U:=1_A - \delta 1_{[1,N]}$. Then there exists a set $S \subset\mathbb{Z}_N$ such that $|S| = O(\delta^{-5})$ and $$\sum_{\xi \in S}|\hat{f}_{U}(\xi)|^2 = \Omega(\delta^2 |S|^{1/5}).$$
Lemma 10.29: Let $N$ and $p$ be as in the above lemma. Let $f:\mathbb{Z}_p \to \mathbb{R}$ be a function supported on $[1,N]$ such that $|f(n)|\le 1$ for all $n$, and such that $$\mathbb{E}_{n \in [1,N]}|f(n)| \le \delta$$ and $$\sum_{\xi \in S} |\hat{f}(\xi)|^2 \ge \sigma$$ for some set $S \subset \mathbb{Z}_p$ and some $\sigma >0$. Then there exists a proper arithmetic progression $P' \subset [1,N]$ with $|P'|=\Omega(N^{1/(|S|+1)})$ and $$\mathbb{E}_{n \in P'}f(n) =\Omega(\sigma /\delta).$$
Here is my work to prove Theorem 10.27:
Now we want to use the above two proposition / lemma to prove the following:
Corollary: Let $A \subset [1,N]$ be such that $|A| = \delta N$ for some $0<\delta\le 1$ and such that $A$ has no proper arithmetic progressions of length $3$. Suppose also that $N \ge 100/(\delta)^2$. Then there exists a set $S \subset\mathbb{Z}_N$ such that $|S| = O(\delta^{-5})$ and there exists a proper arithmetic progression $P' \subset [1,N]$ with $|P'|=\Omega(N^{1/(|S|+1)})$ such that we have a density increment $$P_{P'}(A) \ge P_P(A) +\Omega(\delta |S|^{1/5}).$$
Clearly the corollary follows from Proposition 10.28 and Lemma 10.29 if we choose the function $f$ (in Lemma 10.29) to be $f:=1_A - \delta_{[1,N]}$ as it satisfies all the hypothesis in Lemma 10.29 and so it turns out that $\sigma$ in Lemma 10.29 is same as $\delta^2 |S|^{1/5}$
Now we wish to iterate the corollary to prove Theorem 10.27. So, let the new density is $\delta_{i+1}$ and original density if $\delta_i$, then from corollary it follows that if $N_i$ is large enough ( $N_i \ge 100/(\delta_i)^2$ ) then we can find a subset $S_i$ of $N_i$ such that the new sub-progression has length atleast $\Omega(N_i^{1/(|S_i| +1)}$) and $$\delta_{i+1} \ge \delta_i + \Omega(\delta_i |S_i|^{1/5})$$
Suppose we wish to keep doubling the density , then we can do this atmost $log_2(1/\delta)+1$ as density is always bounded by 1, so the total number of times we can double the density is bounded. Note that the process stops only if $N_i <100/(\delta_i)^2$ . So, if the process terminates at step $i$ then we have : $$N_i < 100/(\delta)_i)^2 \leq 100/(\delta)^2$$ and moreover each time we iterate the corollary we move down to a sub-AP whose length is reduced by a power of $1/(|S_i|+1)$.
Also, the total number of steps $(i)$ that we can keep doubling the density is $O(1/\delta)$. Then we have the following: ($\delta$ is the density of 3-AP free set in $[1,N]$) $$N \le N_i^{{(|S_i|+1)}^i} \le (100/\delta^2)^{{(|S_i|+1)}^{O(1/\delta)}}, \;\;\;\; say (*)$$
But I am unable to conclude from here to show (Theorem 10.27) that $\delta = O(1/\log^c N) \implies |A| = (N/ \log^c N)$. Can someone please tell me where do i go wrong or how to conclude the last step , which shows the density of 3-AP free set. I have explained the steps that I used to reach the conclusion (*) . Please help me to prove this bound. Thanks.