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.)

This entry was posted in Fourier analysis, Number Theory. Bookmark the permalink.

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)!!

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s