Arithmetic progression in some subsets of $\mathbb{Z}_N$ (Part 2)

Last time I have mentioned the idea of the proof (of theorem 2). Now I continue and give a full detail here.

Lemma 7 The Bohr neighbourhood \mathcal{B}(\mathcal{R}, \frac{\eta}{64}) contains an arithmetic progression, named P, of length at least \frac{\eta^3}{2^{17}\log(1/\beta)}N^\frac{\eta^2}{2^{11}\log(1/\beta)}.

Proof. By Lemma 6 (see last post), every element in \mathcal{R} is in the \pm1-linear span of a set \Lambda of size at most m = 2^{11}\eta^{-2}\log(1/\beta). Hence we have \mathcal{B}(\Lambda, \frac{\eta}{64m}) \subset \mathcal{B}(\mathcal{R}, \frac{\eta}{64}) (for, pick x\inLHS, then \|\lambda x\|\leq\frac{\eta}{64m} for all \lambda\in\Lambda, and so by triangle inequality, together with the fact that elements in \mathcal{R} can be written in the form \sum^{m}_{i=1} \epsilon_i \lambda_i, we have \|\sum^{m}_{i=1} \epsilon_i \lambda_i x\|\leq m\|\lambda_i x\|\leq\frac{\eta}{64}.) Note that by Lemma 5, LHS contains an AP of length at least \frac{\eta}{64m}N^\frac{1}{m}. The result follows. \Box

Lemma 8 For at least (1-\frac{\eta}{16})N values of x we have |(x+P)\cap B|\leq\frac{16\beta}{\eta}|P|.

Proof. Suppose not. Then more than \frac{\eta}{16}N values of x we have the reversed inequality. So

\displaystyle \begin{array}{rcl} |P||B| &=& \sum_{x\in\mathbb{Z}_N} |\{(p, b) \in P\times B: p+x = b\}| \text{\quad (Since there are } |P||B| \text{ many choices of } x\text{)} \\ &=& \sum_{x\in\mathbb{Z}_N} |(x+P)\cap B| \\&>& \frac{\eta N}{16}\frac{16\beta}{\eta}|P| \\&=& |P||B|,\end{array}

contradiction. \Box

Now let C = \{x\in\mathbb{Z}_N: |(x+P)\cap B| \leq\frac{16\beta}{\eta}|P|\}. By Lemma 8, |C|\geq (1-\frac{\eta}{16})N. Since the size of C is large, it cannot have huge Fourier coefficient for r\neq 0.
(In fact, for r\neq 0, |\hat{C}(r)| = |\widehat{C^c}(r)|\leq |C^c|\leq \frac{\eta N}{16}\leq \frac{\eta |C|}{8}, where C^c is the complement of C and the first equality is because of the fact \widehat{\mathbb{Z}_N}(r) = 0 whenever r\neq 0.)

We now choose D\subset C, using probabilistic method, with certain size so that the Fourier coefficient of D is small.

Lemma 9 Let t\geq 2^{14}\eta^{-2}\log(N). Also assume t\leq\frac{|C|}{2}. Then there exists D\subset C with size t such that \sup_{r\neq 0} |\hat{D}(r)|\leq\frac{\eta t}{4}.

Proof. Choose a set E\subset C by letting each c\in C be in E with probability t/|C|, these choices being independent.
We first compute the expectations and variances of the Fourier coefficient of E. Recall the notation \omega = e^{\frac{2\pi i}{N}}.

\displaystyle \begin{array}{rcl} \mathbb{E}(\hat{E}(r)) &=& \sum_{x\in C} \mathbb{E}(E(x)\omega^{rx}) \\ &=& \sum_{x\in C} \omega^{rx} p \\ &=& p\hat{C}(r).\end{array}

\displaystyle \begin{array}{rcl} \text{Var}(\hat{E}(r)) &=& \mathbb{E}(|\hat{E}(r)|^2) -|\mathbb{E}(\hat{E}(r))|^2\\ &=& \mathbb{E}(|\sum_{x\in C}E(x)\omega^{rx}|^2)-p^2|\hat{C}(r)|^2\\ &=& \mathbb{E}(\sum_{x\in C}|E(x)|^2+\sum_{x\neq y}E(x)E(y)\omega^{r(x-y)})\\&\phantom{=}&-p^2(\sum_{x\in C}|C(x)|^2+\sum_{x\neq y}|C(x)||C(y)|\omega^{r(x-y)})\\ &=& p|C|+p^2\sum_{x\neq y}\omega^{r(x-y)}-p^2|C|-p^2\sum_{x\neq y}\omega^{r(x-y)}\\ &=& p|C|-p^{2}|C|. \end{array}

Now consider the random variable X^{(r)}_j = (E(j)-p)\omega^{rj}. By above computations we have \mathbb{E}(X^{(r)}_j) = 0, \text{Var}(X^{(r)}_j) = p|C|-p^2|C| \geq \frac{p|C|}{2} = \frac{t}{2}\geq \frac{\eta t}{24}. Also it is easy to see that |X^{(r)}_j| \leq 1. Therefore, we can apply Lemma 4 with these random variables X^{(r)}_j.

Case 1: r\neq0. Then

\displaystyle \begin{array}{rcl} \mathbb{P}(|\hat{E}(r)|\geq\frac{\eta t}{6}) &\leq&\mathbb{P}(|\hat{E}(r)-\mathbb{E}(\hat{E}(r))|\geq\frac{\eta t}{24})\text{\qquad (Triangle inequality)}\\&<&4e^{-\frac{\eta^2t^2}{8\times24^2(p|C|-p^2|C|)}}\text{\qquad (Lemma 4)}\\&<&4e^{-\frac{\eta^2t}{5000}}\end{array}

Case 2: r=0. Then \displaystyle\mathbb{P}(||E|-t|\geq\frac{\eta t}{24})<4e^{-\frac{\eta^2t}{5000}} by Lemma 4.

Therefore, if t \geq 2^{14}\eta^{-2}\log{N}, 4e^{-\frac{\eta^2t}{5000}} < 1/2, and so there is a positive probability such that ||E|-t| < \frac{\eta t}{24} and |\hat{E}(r)|<\frac{\eta t}{6} happens. By adding or deleting suitable number of elements (which is at most \frac{\eta t}{24})  from E we will get D (Since the number of changed elements is at most \frac{\eta t}{24}, |\hat{D}(r)-\hat{E}(r)| \leq \frac{\eta t}{24} and so |\hat{D}(r)|\leq \frac{\eta t}{4}). \Box

With the same proof we get the following lemma:

Lemma 10 Let t\geq 2^{14}\eta^{-2}\log(N). Also assume t\leq\beta N/2. Then there exists X\subset B with size t such that |\hat{X}(r) - \frac{t\hat{B}(r)}{|B|}|\leq\frac{\eta t}{12} for r\neq0.

Remark: The above two lemmas actually mean that for any set G\subset \mathbb{Z}_N, there is always a subset H such that 1. |G|/2\geq|H|\geq 2^{14}\eta^{-2}\log{N}; 2. H has Fourier coefficients close to \frac{|H|}{|G|}\hat{G}(r).

Lemma 11 Let S = (B\setminus X)\cup D. Assume (B\setminus X)\cap D=\emptyset. Then \sup_{r\in\mathcal{R}}|\hat{S}(r)|\leq\eta S - \frac{\eta t}{6} and |\hat{S}(r)|\leq \frac{\eta|S|}{2} + \frac{\eta t}{3} for r\in\mathcal{R}.

(Little remark: I avoid the notion “multiset” which is used in original paper. Although the concept of multiset is easy to understand but somehow there is ambiguity, and so I prefer saying that S is disjoint union of two sets instead of saying S is a multiset.)
Proof. Note that
\displaystyle \begin{array}{rcl} \hat{S}(r)&=&\hat{B}(r)-\hat{X}(r)+\hat{D}(r)\\&=&(1-\frac{t}{|B|})\hat{B}(r)+(\hat{D}(r)-(\hat{X}(r)-\frac{t\hat{B}(r)}{|B|})).\end{array}
If r\in\mathcal{R},

\displaystyle \begin{array}{rcl} |\hat{S}(r)|&\leq&(1-\frac{t}{|B|})|\hat{B}(r)|+\frac{\eta t}{3}\text{\qquad (By above calculation, lemma 9 and lemma 10)}\\&\leq&|\hat{B}(r)|-\frac{\eta t}{2}+\frac{\eta t}{3}\text{\qquad (Definition of }\mathcal{R}\text{)}\\&\leq&\eta|B|-\frac{\eta t}{6}\\&=&\eta|S|-\frac{\eta t}{6}.\text{\qquad (Since }|S|=|B|\text{)}\end{array}

If r\not\in\mathcal{R},

\displaystyle \begin{array}{rcl} |\hat{S}(r)|&\leq&(1-\frac{t}{|B|})|\hat{B}(r)|+\frac{\eta t}{3}\\&\leq&\frac{\eta|B|}{2}+\frac{\eta t}{3}\text{\qquad (Again definition of }\mathcal{R}\text{)}\\&=&\frac{\eta|S|}{2}+\frac{\eta t}{3}.\end{array}

The result follows. \Box

Write D=\{d_1, \cdots, d_t\}. Let D' be any set obtained by replacing d_j by x_j+d_j, where x_j\in P. (Recall: P is an AP in the Bohr neighbourhood of \mathcal{R})

Lemma 12 Suppose that t\leq\frac{\eta\beta N}{10}. Let S' = (B\setminus X)\cup D'. Assume the union is disjoint. Then \sup_{r\neq0} |\widehat{S'}(r)| < \eta |S'|.

Proof. If r\not\in\mathcal{R}, as we changed the elements in D, |\widehat{S'}(r)-\hat{S}(r)| \leq \sum_{x\in D'} + \sum_{x\in D} 1 = 2t. Hence

\displaystyle \begin{array}{rcl} |\widehat{S'}(r)|&\leq&2t+|\hat{S}(r)|\\&\leq&\frac{\eta|S'|}{2}+\frac{\eta t}{2}+2t\\&<&\frac{\eta|S'|}{2}+5t\\&\leq&\frac{\eta|S'|}{2}+\frac{\eta|S'|}{2}\\&=&\eta|S|.\end{array}

If r\in\mathcal{R}, recall P\subset\mathcal{B}(\mathcal{R}, \frac{\eta}{64}), then

\displaystyle \begin{array}{rcl} |\widehat{S'}(r)|&\leq&2t+|\hat{S}(r)|\\&\leq&\sum_{j=1}^{t} |\omega^{r(d_j+x_j)}-\omega^{rd_j}|\\&\leq&t\sup_{j}|\omega^{rx_j}-1|\\&\leq&\frac{\eta t}{8}.\text{\qquad(Since }\|rx_j\|\leq\frac{\eta}{64}\text{ together with some elementary inequalities)}\end{array}

Therefore |\widehat{S'}(r)|\leq\frac{\eta t}{8}+|\hat{S}(r)| = \frac{\eta t}{8}+\eta|S'|-\frac{\eta t}{6} < \eta|S'|. \Box

We are almost done (finally!). Recall B\subset A is the set such that \sup_{r\neq0}|\hat{B}(r)| is minimal subject to the constraint |B| = \beta N. Therefore if D'\subset A, S' could not exist! In another words, no such choice of x_1,\cdots, x_t such that one could construct S'.

Lemma 13 There exists some j such that d_j + P is contained in B\cup A^c, except for at most t elements, which could lie in A\setminus B.

Proof. Suppose not. Then for all j, more than t elements in d_j+P lie in A\setminus B. Choose x_1\in P such that d_1+x_1\in A\setminus B. Choose x_2\in P such that d_2+x_2 \in A\setminus (B\cup\{d_1+x_1\}). Continue this way, choose x_t\in P such that d_t+x_t\in A\setminus (B\cup \bigcup_{j=1}^{t-1}\{d_j+x_j\}). This gives us a D'\subset A, contradiction. \Box

Let j be such that Lemma 13 holds. Recall that d_j\in C. Therefore |(d_j+P)\cap B|\leq\frac{16\beta}{\eta}|P|. Therefore, A^c contains at least (1-\frac{16\beta}{\eta})|d_j+P| and minus at most t points.
If one choose t\leq\frac{16\beta}{\eta}|P|, A^c will contain a portion (1-\frac{32\beta}{\eta}) of d_j+P. Suppose further that the parameters have also been chosen such that |P|\geq\frac{\eta}{32\beta}. Then A^c contains more than (1-\frac{32\beta}{\eta}) of P must contain an AP of length at least \frac{\eta}{96\beta}.

Remark: If we choose \beta such that |A|<\beta N, then it is even impossible to define B. But in this case A^c will contain an AP of length at least 1/\beta > \eta/96\beta.

And hence above leads to

Proposition 14 Suppose that t\leq\frac{16\beta}{\eta}|P|, |P|\geq\frac{\eta}{32\beta}, and t satisfies the condition in Lemma 9, 10. Then A^c contains an AP of length at least \frac{\eta}{96\beta}.

By taking \beta=e^{-\alpha\sqrt{\log{N}}/16}, t = 2^{14}\eta^{-2}\log{N}, with some painful calculations we can get theorem 2. (Actually I didn’t check this, it may be wrong. But I think this part is less important compared to the whole argument.)

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One Response to Arithmetic progression in some subsets of $\mathbb{Z}_N$ (Part 2)

  1. wilson says:

    It is not so much for me (2005 MSc HKBU and 2010 MSc PolyU)!!

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